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✓ 35 results · all verified · 23 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 12 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Winding Number and the Global Cauchy Theorem

1 · Prerequisites

2 · Summary

This page starts from the local complex tools already available on discs and contours: complex line integrals, the ML estimate, the one-variable Cauchy formula, local holomorphic logarithms of nonvanishing functions, and the topological facts that open plane sets decompose into connected components and that compact plane sets have a unique unbounded complementary component. Those inputs are enough to define winding numbers for rectifiable closed contours without assuming differentiability of the parameter.

The development first builds continuous logarithms along a contour and uses them to prove that the winding number is integral, locally constant off the trace, and zero on the unbounded complementary component. It then extends the index and integration from one contour to finite chains and cycles, introduces null homology and homological simple connectivity, and proves the global Cauchy integral formula, Cauchy's theorem for null-homologous cycles, the invariance of holomorphic integrals under homology, and the existence of primitives, holomorphic logarithms, and holomorphic roots on homologically simply connected domains.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Tagged sums approximate a contour integral within oscillation times length

Statement

Let $γ : [a, b] \to C$ be a rectifiable contour with $a < b$ , let $f$ be continuous on its trace $γ^{*}$ , let $a = t_{0} < t_{1} < \dots < t_{r} = b$ be a partition of $[a, b]$ , and choose a tag $ξ_{i} \in [t_{i}, t_{i + 1}]$ for each $i < r$ . Write $γ_{i}$ for the restriction $γ ∣_{[t_{i}, t_{i + 1}]}$ and

$ω_{i} : = \sup {∣ f (u) - f (v) ∣ : u, v \in γ ([t_{i}, t_{i + 1}])},$

which is a nonnegative real number. Then

$∣ \int_{γ} f (z) d z - \sum_{i < r} f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i})) ∣ \leq \sum_{i < r} ω_{i} L (γ_{i}) .$

In particular, if $ω \geq 0$ satisfies $∣ f (u) - f (v) ∣ \leq ω$ for all $u, v \in γ^{*}$ , then the left-hand side is at most $ω L (γ)$ .

The bound is stated with the oscillations themselves and not as a limit, so a modulus of continuity for $f$ on $γ^{*}$ converts directly into an error estimate. For a singleton parameter interval $[a, a]$ there is no partition, and both the integral and the empty tagged sum are $0$ .

Facts & Assumptions

Given: A rectifiable contour $γ : [a, b] \to C$ with $a < b$ , a continuous $f$ on $γ^{*}$ , a partition $a = t_{0} < t_{1} < \dots < t_{r} = b$ , and tags $ξ_{i} \in [t_{i}, t_{i + 1}]$ for $i < r$ .

[L1]

A complex contour is a rectifiable path $γ : [a, b] \to C$ ; if $α, β : [0, 1] \to C$ satisfy $α (1) = β (0)$ , their concatenation is $(α * β) (s) = α (2 s)$ for $0 \leq s \leq \frac{1}{2}$ and $β (2 s - 1)$ for $\frac{1}{2} \leq s \leq 1$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L2]

For a rectifiable $γ : [a, b] \to C$ and $f$ continuous on its trace, the complex line integral $\int_{γ} f d z$ of The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L3]

If $ϕ : [c, d] \to [a, b]$ is a strictly increasing continuous bijection, $γ : [a, b] \to C$ is rectifiable and $f$ is continuous on the trace of $γ$ , then $\int_{γ \circ ϕ} f d z = \int_{γ} f d z$ (Complex and absolute line integrals are invariant under increasing continuous reparametrization).

[L4]

For composable rectifiable contours $α, β$ , $\int_{α * β} f d z = \int_{α} f d z + \int_{β} f d z$ (Complex line integrals change sign under reversal and add under concatenation).

[L5]

For continuous $f, g$ on the trace of a rectifiable contour $γ$ and $α, β \in C$ , $\int_{γ} (α f + β g) d z = α \int_{γ} f d z + β \int_{γ} g d z$ (Complex line integrals are linear in the integrand).

[L6]

For $c \in C$ and a rectifiable contour $γ : [a, b] \to C$ , $\int_{γ} c d z = c (γ (b) - γ (a))$ (The contour integral of a constant c is c times the endpoint displacement).

[L7]

If $∣ f (z) ∣ \leq M$ on the trace of a rectifiable contour $γ$ , with $M \geq 0$ , then $∣ \int_{γ} f (z) d z ∣ \leq M L (γ)$ (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L8]

For a path $γ : [a, b] \to R^{n}$ with $n \geq 1$ and $c \in [a, b]$ , $L_{[a, b]} (γ) = L_{[a, c]} (γ ∣_{[a, c]}) + L_{[c, b]} (γ ∣_{[c, b]})$ in the nonnegative extended reals, and $γ$ is rectifiable on $[a, b]$ if and only if both restrictions are rectifiable (Arc length is additive across every subdivision point and decreases under restriction).

[L9]

A partition of $[a, b]$ with $a < b$ consists of $a = t_{0} < t_{1} < \dots < t_{r} = b$ with $r \geq 1$ , its subintervals $[t_{i}, t_{i + 1}]$ being indexed from $i = 0$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L10]

$∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for all $z, w \in C$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L11]

If $a_{k} \leq b_{k}$ for all $k < n$ then $\sum_{k < n} a_{k} \leq \sum_{k < n} b_{k}$ (Laws of finite sums and finite products).

[L12]

If a property holds at $0$ and passes from $n$ to $n + 1$ , it holds for every $n \in N$ (The principle of mathematical induction).

[L13]

A closed box in $R^{n}$ is compact, and a subset of $R^{n}$ is compact exactly when it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L14]

The continuous image of a compact subset is a compact subset (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L15]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

Proof

technique · direct

1.1givenL8L9L12

By [L8] applied at $t_{1}$ , then to $γ ∣_{[t_{1}, b]}$ at $t_{2}$ , and so on, an induction on the number of partition points ([L12]) shows that each $γ_{i}$ is rectifiable and that $L (γ) = \sum_{i < r} L (γ_{i})$ .

1.2givenL13L14L15

Each $ω_{i}$ is a nonnegative real: $[t_{i}, t_{i + 1}]$ is a closed bounded interval, hence compact by [L13]; $f \circ γ$ is continuous on it, so its image is compact by [L14] and bounded by [L15]; hence ${∣ f (u) - f (v) ∣ : u, v \in γ ([t_{i}, t_{i + 1}])}$ is a nonempty set of reals bounded above, and it has a supremum, which is $\geq 0$ because $u = v$ is allowed.

1.3givenL1L2L3L4

For $a \leq u < v < w \leq b$ put $α (s) = γ (u + s (v - u))$ and $β (s) = γ (v + s (w - v))$ on $[0, 1]$ ; then $α (1) = γ (v) = β (0)$ , so $α * β$ is defined by [L1], and $α * β = γ ∣_{[u, w]} \circ ϕ$ where $ϕ : [0, 1] \to [u, w]$ is the strictly increasing continuous bijection that is affine on $[0, \frac{1}{2}]$ and on $[\frac{1}{2}, 1]$ with $ϕ (\frac{1}{2}) = v$ . Since $α$ and $β$ are increasing reparametrisations of $γ ∣_{[u, v]}$ and $γ ∣_{[v, w]}$ , [L3] and [L4] give $\int_{γ ∣_{[u, w]}} f d z = \int_{γ ∣_{[u, v]}} f d z + \int_{γ ∣_{[v, w]}} f d z$ .

1.4L6

For each $i < r$ , [L6] applied to the constant $f (γ (ξ_{i}))$ on $γ_{i}$ gives $\int_{γ_{i}} f (γ (ξ_{i})) d z = f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i}))$ .

2.1step 1.3L12

Applying step 1.3 at $t_{1}$ , then to $γ ∣_{[t_{1}, b]}$ at $t_{2}$ , and so on, an induction on the number of partition points ([L12]) gives $\int_{γ} f d z = \sum_{i < r} \int_{γ_{i}} f d z$ .

2.2step 1.2step 1.4L5L7

Fix $i < r$ . The tag value $γ (ξ_{i})$ lies on the trace of $γ_{i}$ , so $∣ f (z) - f (γ (ξ_{i})) ∣ \leq ω_{i}$ for every $z$ on that trace by the definition of $ω_{i}$ in step 1.2; by [L5] the difference $\int_{γ_{i}} f d z - \int_{γ_{i}} f (γ (ξ_{i})) d z$ equals $\int_{γ_{i}} (f (z) - f (γ (ξ_{i}))) d z$ , and [L7] bounds its modulus by $ω_{i} L (γ_{i})$ .

3.1step 2.1step 2.2L10L11L12

Subtracting the identity of step 1.4 from that of step 2.1 termwise, the quantity to be estimated is $\sum_{i < r} (\int_{γ_{i}} f d z - f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i})))$ ; the finite triangle inequality, obtained from [L10] by induction ([L12]), and then [L11] with the bounds of step 2.2, give the stated estimate $\sum_{i < r} ω_{i} L (γ_{i})$ .

4.1step 1.1step 3.1L11∎

If $∣ f (u) - f (v) ∣ \leq ω$ for all $u, v \in γ^{*}$ then $ω_{i} \leq ω$ for every $i < r$ , so step 3.1 and [L11] bound the error by $ω \sum_{i < r} L (γ_{i})$ , which is $ω L (γ)$ by step 1.1; and on a singleton interval $[a, a]$ the integral is $0$ and there is no partition, so the assertion made there is the stated one about the empty sum.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A continuous function holomorphic off a single point is holomorphic

Statement

Let $U \subseteq C$ be open, let $p \in U$ , and let $f : U \to C$ be continuous on $U$ and holomorphic on $U ∖ {p}$ . Then $f$ is holomorphic on $U$ , the point $p$ included.

Facts & Assumptions

Given: An open set $U \subseteq C$ , a point $p \in U$ , and a function $f : U \to C$ that is continuous on $U$ and holomorphic on $U ∖ {p}$ .

[L1]

If $U \subseteq C$ is open, $p \in U$ , and $f : U \to C$ is continuous and holomorphic on $U ∖ {p}$ , then $\int_{\partial T} f (z) d z = 0$ for every filled triangle $T = Δ [a, b, c]$ of Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter contained in $U$ ; the exceptional point may lie outside, inside, or on the boundary of $T$ (Goursat's triangle theorem remains valid for a continuous function holomorphic away from one point).

[L2]

If $Ω \subseteq C$ is open and $f : Ω \to C$ is continuous, then $f$ is holomorphic on $Ω$ if and only if $\int_{\partial Δ [a, b, c]} f (z) d z = 0$ whenever $Δ [a, b, c] \subseteq Ω$ ; repeated or collinear vertices are permitted (Morera's theorem: vanishing triangle integrals characterize holomorphy among continuous functions).

Proof

technique · direct

1.1givenL1

The hypotheses of [L1] are exactly the given ones, so $\int_{\partial T} f (z) d z = 0$ for every filled triangle $T \subseteq U$ , whether $p$ lies outside $T$ , inside it, or on its boundary.

2.1givenstep 1.1L2∎

The function $f$ is continuous on the open set $U$ and step 1.1 supplies the vanishing triangle integrals demanded by the right-hand side of [L2], so [L2] makes $f$ holomorphic on all of $U$ , including at $p$ .

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-26Open item page →

A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic

Statement

Let $γ : [a, b] \to C$ be a rectifiable contour with trace $γ^{*}$ , let $Ω \subseteq C$ be open, and let $φ : γ^{*} \times Ω \to C$ be continuous, with $z \mapsto φ (w, z)$ holomorphic on $Ω$ for every $w \in γ^{*}$ . Then

$F (z) : = \int_{γ} φ (ζ, z) d ζ$

is defined for every $z \in Ω$ and is holomorphic on $Ω$ .

Here $γ^{*} \times Ω$ carries the Euclidean metric of $R^{4}$ under the coordinate identification of the plane, so continuity of $φ$ is joint continuity in the two variables together.

Facts & Assumptions

Given: A rectifiable contour $γ : [a, b] \to C$ , an open $Ω \subseteq C$ , and a continuous $φ : γ^{*} \times Ω \to C$ with $φ (w, \cdot)$ holomorphic on $Ω$ for each $w \in γ^{*}$ ; products of subsets of $C$ are read in $R^{4}$ through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves, and locally uniform convergence is that of Locally uniform convergence on an open subset of the complex plane is compact convergence.

[L1]

For a rectifiable contour $γ : [a, b] \to C$ with $a < b$ , a continuous $f$ on $γ^{*}$ , a partition $a = t_{0} < \dots < t_{r} = b$ and tags $ξ_{i} \in [t_{i}, t_{i + 1}]$ , the difference between $\int_{γ} f d z$ and $\sum_{i < r} f (γ (ξ_{i})) (γ (t_{i + 1}) - γ (t_{i}))$ has modulus at most $ω L (γ)$ whenever $ω \geq 0$ satisfies $∣ f (u) - f (v) ∣ \leq ω$ for all $u, v \in γ^{*}$ (Tagged sums approximate a contour integral within oscillation times length).

[L2]

If each $f_{n} : Ω \to C$ is holomorphic on an open $Ω$ and $f_{n} \to f$ locally uniformly on $Ω$ , then $f$ is holomorphic (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[L3]

A continuous map from a compact metric space to a metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L4]

A subset of $R^{n}$ is compact exactly when it is closed and bounded, and every closed box is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L5]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

[L6]

The continuous image of a compact subset is a compact subset (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L7]

A linear combination $α f + β g$ of functions complex differentiable at a point is complex differentiable there, with $(α f + β g)^{'} = α f^{'} + β g^{'}$ , and every constant function has derivative $0$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L8]

For a rectifiable $γ : [a, b] \to C$ and $f$ continuous on its trace, the complex line integral $\int_{γ} f d z$ exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L9]

For $a < b$ and a natural $N \geq 1$ the uniform partition of $[a, b]$ into $N$ parts has points $t_{i} = a + i (b - a) / N$ and mesh $(b - a) / N$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L10]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L11]

$B (x, r)$ is the set of points at distance below $r$ from $x$ and $\overset{ˉ}{B} (x, r)$ the set at distance at most $r$ ; a set is open exactly when each of its points has some ball around it inside the set (Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L12]

A complex contour is a rectifiable path $γ : [a, b] \to C$ , so $L (γ)$ is a nonnegative real (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

Proof

technique · direct

1.1givenL4L6

The parameter interval $[a, b]$ is compact by [L4] and $γ$ is continuous, so the trace $γ^{*}$ is compact by [L6].

1.2givenL8

For each $z \in Ω$ the map $ζ \mapsto φ (ζ, z)$ is continuous on $γ^{*}$ , so $F (z)$ exists by [L8].

1.3givenL7L9

Write $P_{N}$ for the uniform partition of $[a, b]$ into $N$ parts, with points $t_{i}^{N} = a + i (b - a) / N$ , and set $S_{N} (z) = \sum_{i < N} φ (γ (t_{i}^{N}), z) (γ (t_{i + 1}^{N}) - γ (t_{i}^{N}))$ for $z \in Ω$ , assuming $a < b$ . Each summand is a constant multiple of a function holomorphic on $Ω$ , so $S_{N}$ is holomorphic on $Ω$ by [L7].

2.1step 1.1L3L4L5L11

Fix $z_{0} \in Ω$ . By [L11] there is $ρ > 0$ with $\overset{ˉ}{B} (z_{0}, ρ) \subseteq Ω$ ; put $K = \overset{ˉ}{B} (z_{0}, ρ)$ , which is closed and bounded, hence compact by [L4]. By [L5] and step 1.1 both $γ^{*}$ and $K$ are closed and bounded, so $γ^{*} \times K$ is a closed bounded subset of $R^{4}$ and is compact by [L4]; since $φ$ is continuous there, [L3] makes it uniformly continuous on $γ^{*} \times K$ .

3.1step 2.1L3L4L10choose

Let $ε > 0$ . Step 2.1 gives $η > 0$ such that $∣ φ (u, z) - φ (v, z) ∣ \leq ε$ whenever $u, v \in γ^{*}$ satisfy $∣ u - v ∣ < η$ and $z \in K$ . The interval $[a, b]$ is compact by [L4], so $γ$ is uniformly continuous on it by [L3]: there is $δ > 0$ with $∣ γ (t) - γ (s) ∣ < η$ whenever $∣ t - s ∣ < δ$ . By [L10] applied to $δ / (b - a)$ there is a natural $n \geq 1$ with $(b - a) / n < δ$ .

4.1step 1.2step 1.3step 3.1L1L12

Let $N \geq n$ and $z \in K$ . Every two parameters in a subinterval of $P_{N}$ differ by at most $(b - a) / N \leq (b - a) / n < δ$ , so any two points of $γ ([t_{i}^{N}, t_{i + 1}^{N}])$ are within $η$ of each other and step 3.1 bounds $∣ φ (u, z) - φ (v, z) ∣$ by $ε$ for such points. Applying [L1] to $f = φ (\cdot, z)$ on each subarc, with $ω = ε$ on that subarc, and summing the subarc bounds gives $∣ F (z) - S_{N} (z) ∣ \leq ε L (γ)$ .

5.1step 1.3step 4.1L2L7L11L12∎

Since $ε > 0$ was arbitrary and $L (γ)$ is a fixed nonnegative real by [L12], step 4.1 says $S_{N} \to F$ uniformly on $K$ , hence uniformly on the open neighbourhood $B (z_{0}, ρ)$ of $z_{0}$ ; as $z_{0} \in Ω$ was arbitrary, $S_{N} \to F$ locally uniformly on $Ω$ , and [L2] with step 1.3 makes $F$ holomorphic on $Ω$ . If instead $a = b$ then $\int_{γ} f d z = 0$ for every continuous $f$ , so $F$ is identically $0$ and holomorphic by [L7].

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A holomorphic logarithm is a primitive of the logarithmic derivative

Statement

Let $U \subseteq C$ be open and let $L, h : U \to C$ be holomorphic with $\exp (L (z)) = h (z)$ for every $z \in U$ . Then $h$ is nowhere zero on $U$ and

$L^{'} (z) = \frac{h^{'} (z)}{h (z)} (z \in U) .$

In particular, if $p \in C$ , if $U$ misses $p$ , and if $L$ is holomorphic on $U$ with $\exp (L (z)) = z - p$ for every $z \in U$ , then $L^{'} (z) = 1 / (z - p)$ on $U$ .

Facts & Assumptions

Given: An open $U \subseteq C$ and holomorphic $L, h : U \to C$ with $\exp \circ L = h$ .

[L1]

The complex exponential is entire and $\exp^{'} (z) = \exp z$ for every $z \in C$ (The complex exponential is entire and its complex derivative is itself).

[L2]

If $f : U \to V$ is complex differentiable at $a$ and $g : V \to C$ is complex differentiable at $f (a)$ , then $(g \circ f)^{'} (a) = g^{'} (f (a)) f^{'} (a)$ (The chain rule for complex derivatives).

[L3]

Linear combinations and products of functions complex differentiable at a point are complex differentiable there, with the usual formulas; if $g (a) \neq 0$ then $(f / g)^{'} (a) = (f^{'} (a) g (a) - f (a) g^{'} (a)) / g (a)^{2}$ ; every constant function has derivative $0$ and the identity function has derivative $1$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L4]

For real $x, y$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ and $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L5]

A function is holomorphic on an open $U$ when it is complex differentiable at every point of $U$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

Proof

technique · direct

1.1givenL4

For every $v \in C$ , writing $v = x + i y$ with $x, y$ real, [L4] gives $∣ \exp v ∣ = e^{x} > 0$ , so $\exp$ never vanishes; hence $h = \exp \circ L$ is nowhere zero on $U$ .

1.2givenL1L2L5

By [L5] both $L$ and $h$ are complex differentiable at every point of $U$ , and [L1] and [L2] give $(\exp \circ L)^{'} (z) = \exp (L (z)) L^{'} (z)$ there.

2.1givenstep 1.1step 1.2algebra

Since $\exp \circ L = h$ as functions on $U$ , step 1.2 says $h^{'} (z) = h (z) L^{'} (z)$ for every $z \in U$ ; dividing by the nonzero $h (z)$ of step 1.1 gives $L^{'} (z) = h^{'} (z) / h (z)$ .

3.1step 2.1L3∎

If $U$ misses $p$ and $h (z) = z - p$ on $U$ , then $h^{'} (z) = 1$ by [L3], so step 2.1 gives $L^{'} (z) = 1 / (z - p)$ on $U$ .

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A disc missing $p$ carries a holomorphic logarithm of $z - p$

Statement

Let $D (c, ρ) = B (c, ρ)$ be an open disc in $C$ with $ρ > 0$ and let $p \in C$ with $p \notin D (c, ρ)$ . Then there is a holomorphic $L : D (c, ρ) \to C$ with

$\exp (L (z)) = z - p for every z \in D (c, ρ),$

and every such $L$ satisfies $L^{'} (z) = 1 / (z - p)$ there. If $L_{1}$ and $L_{2}$ both have this property, then $L_{1} - L_{2}$ is a constant lying in $2 π i Z$ .

Facts & Assumptions

Given: An open disc $D (c, ρ)$ with $ρ > 0$ and a point $p \notin D (c, ρ)$ .

[L1]

If $D (a, r)$ is an open disc with $r > 0$ and $h : D (a, r) \to C$ is holomorphic and nowhere zero, then there is a holomorphic $L : D (a, r) \to C$ with $\exp (L (z)) = h (z)$ for every $z \in D (a, r)$ (A nonvanishing holomorphic function on a disc has a holomorphic logarithm).

[L2]

If $L$ and $h$ are holomorphic on an open $U$ with $\exp \circ L = h$ , then $h$ is nowhere zero and $L^{'} = h^{'} / h$ ; if $U$ misses $p$ and $h (z) = z - p$ , then $L^{'} (z) = 1 / (z - p)$ (A holomorphic logarithm is a primitive of the logarithmic derivative).

[L3]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L4]

The continuous image of a connected subset is a connected subset (A continuous image of a connected space is connected, and connectedness is a topological property).

[L5]

A subset $E \subseteq R$ is connected exactly when it is order-convex: $x, z \in E$ and $x \leq w \leq z$ imply $w \in E$ (The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ ").

[L6]

A path-connected subset of a topological space is a connected subset (Every path-connected space is connected, and every path component lies inside a component); a subset is path-connected when any two of its points are joined by a continuous map from $[0, 1]$ with image inside it (Paths, path-connected spaces and path components).

[L7]

A subset $U \subseteq R^{m}$ is convex when $(1 - t) x + t y \in U$ for all $x, y \in U$ and $t \in [0, 1]$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L8]

$B (x, r) = {y : d (x, y) < r}$ (Open ball, closed ball and sphere in a metric space).

[L9]

Linear combinations, products and nonvanishing quotients of functions complex differentiable at a point are complex differentiable there; constants have derivative $0$ and the identity has derivative $1$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L10]

A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L11]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for complex $z, w$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L12]

For $z = a + b i$ with $a, b$ real, $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

[L13]

The integers form an ordered commutative ring, and their canonical image in $R$ is discrete; hence if $m < n$ then $m + \frac{1}{2}$ lies strictly between them and is not an integer (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

Proof

technique · direct

1.1givenL1L2L9

The function $h (z) = z - p$ is holomorphic on $D (c, ρ)$ by [L9], and it is nowhere zero there because $p \notin D (c, ρ)$ ; so [L1] supplies a holomorphic $L$ on $D (c, ρ)$ with $\exp \circ L = h$ , and [L2] gives $L^{'} (z) = 1 / (z - p)$ for every such $L$ .

1.2L6L7L8L11

$D (c, ρ)$ is convex in the sense of [L7]: for $z, w$ in it and $t \in [0, 1]$ , [L8] and [L11] give $∣ (1 - t) z + t w - c ∣ = ∣ (1 - t) (z - c) + t (w - c) ∣ \leq (1 - t) ∣ z - c ∣ + t ∣ w - c ∣ < ρ$ . Hence any two of its points are joined by the continuous map $t \mapsto (1 - t) z + t w$ of $[0, 1]$ into it, so $D (c, ρ)$ is path-connected and therefore a connected subset of $C$ by [L6].

2.1step 1.1L3L9L10L12

Let $L_{1}, L_{2}$ both be holomorphic on $D (c, ρ)$ with $\exp \circ L_{j} = h$ . Then $\exp (L_{1} (z)) = \exp (L_{2} (z))$ for every $z$ , so $L_{1} (z) - L_{2} (z) \in 2 π i Z$ by [L3]; in particular $Re (L_{1} - L_{2}) = 0$ and the function $g : = Im (L_{1} - L_{2}) / (2 π)$ takes values in $Z$ . By [L9] and [L10] the difference $L_{1} - L_{2}$ is continuous, and $∣ Im v ∣ \leq ∣ v ∣$ by [L12], so $g$ is a continuous real-valued function on $D (c, ρ)$ .

3.1step 1.2step 2.1L4L5L13∎

By step 1.2 and [L4] the image $g [D (c, ρ)]$ is a connected subset of $R$ , hence order-convex by [L5]; if it contained two distinct integers $m < n$ it would contain $m + \frac{1}{2}$ , which is not an integer, contradicting step 2.1 and [L13]. So $g$ is constant, and $L_{1} - L_{2}$ is the constant $2 π i g \in 2 π i Z$ .

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A contour missing a point subdivides into arcs lying in discs that miss it

Statement

Let $γ : [a, b] \to C$ be a complex contour with trace $γ^{*}$ and let $p \in C$ with $p \notin γ^{*}$ . Then

$d : = \inf {∣ w - p ∣ : w \in γ^{*}}$

exists and satisfies $d > 0$ , and there is $δ > 0$ with the following property: whenever $a < b$ and $a = t_{0} < t_{1} < \dots < t_{r} = b$ is a partition of $[a, b]$ of mesh smaller than $δ$ ,

$γ ([t_{i}, t_{i + 1}]) \subseteq D (γ (t_{i}), d) and p \notin D (γ (t_{i}), d) for every i < r,$

where $D (u, d)$ is the open disc of centre $u$ and radius $d$ . At least one such partition exists. If instead $a = b$ the trace is the single point $γ (a)$ , which lies in $D (γ (a), d)$ , and $p \notin D (γ (a), d)$ ; no partition is involved in that case.

Facts & Assumptions

Given: A complex contour $γ : [a, b] \to C$ and a point $p \notin γ^{*}$ ; the plane carries the Euclidean metric of $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

A complex contour is a rectifiable path $γ : [a, b] \to C$ , in particular a continuous map on a compact interval (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

[L2]

The continuous image of a compact subset is a compact subset (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L3]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

[L4]

A continuous map from a compact metric space to a metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L5]

A subset of $R^{n}$ is compact exactly when it is closed and bounded, and closed boxes are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L6]

$B (x, r) = {y : d (x, y) < r}$ , and a set is open exactly when each of its points admits a ball around it inside the set, a set being closed when its complement is open (Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L7]

A partition of $[a, b]$ with $a < b$ consists of $a = t_{0} < \dots < t_{r} = b$ with $r \geq 1$ ; its mesh is the largest of the lengths $t_{i + 1} - t_{i}$ , and the uniform partition into $N$ parts has mesh $(b - a) / N$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L8]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L9]

A nonempty subset of $R$ bounded below has a greatest lower bound (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

Proof

technique · direct

1.1givenL1L2L3L5

By [L1] and [L5] the parameter interval is compact and $γ$ is continuous, so $γ^{*}$ is a nonempty compact subset of $C$ by [L2], and it is closed by [L3].

1.2givenL1L4L5

By [L1] and [L5] again, $γ$ is uniformly continuous on $[a, b]$ by [L4].

2.1step 1.1L6L9

The set ${∣ w - p ∣ : w \in γ^{*}}$ is nonempty and bounded below by $0$ , so $d$ exists by [L9]. Since $p \notin γ^{*}$ and $γ^{*}$ is closed by step 1.1, its complement is open, so [L6] gives $ε > 0$ with $B (p, ε) \cap γ^{*} = \emptyset$ , that is $∣ w - p ∣ \geq ε$ for every $w \in γ^{*}$ ; hence $d \geq ε > 0$ .

3.1step 1.2step 2.1choose

Apply the uniform continuity of step 1.2 with the positive number $d$ of step 2.1: there is $δ > 0$ such that $∣ γ (t) - γ (s) ∣ < d$ whenever $s, t \in [a, b]$ satisfy $∣ t - s ∣ < δ$ .

4.1step 2.1step 3.1L6L7

Let $a < b$ and let $a = t_{0} < \dots < t_{r} = b$ have mesh below $δ$ . For $i < r$ and $t \in [t_{i}, t_{i + 1}]$ one has $∣ t - t_{i} ∣ \leq t_{i + 1} - t_{i} < δ$ , so $∣ γ (t) - γ (t_{i}) ∣ < d$ by step 3.1 and hence $γ (t) \in D (γ (t_{i}), d)$ by [L6]; and $∣ p - γ (t_{i}) ∣ \geq d$ by step 2.1, since $γ (t_{i}) \in γ^{*}$ , so $p \notin D (γ (t_{i}), d)$ .

5.1step 2.1step 4.1L6L7L8∎

Such a partition exists when $a < b$ : by [L8] applied to $δ / (b - a)$ there is a natural $N \geq 1$ with $(b - a) / N < δ$ , and the uniform partition into $N$ parts has mesh $(b - a) / N < δ$ by [L7]. If $a = b$ then $γ^{*} = {γ (a)}$ , which lies in $D (γ (a), d)$ because $d > 0$ , while $∣ p - γ (a) ∣ \geq d$ keeps $p$ out of that disc.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Continuous logarithms and continuous arguments along a contour

Definition

Let $γ : [a, b] \to C$ be a complex contour (Rectifiable complex contours, reversal, concatenation, closedness, and orientation) with trace $γ^{*}$ , and let $p \in C$ with $p \notin γ^{*}$ .

A continuous logarithm of $γ - p$ along $γ$ is a continuous function $λ : [a, b] \to C$ with

$\exp (λ (t)) = γ (t) - p for every t \in [a, b],$

the exponential being that of The complex exponential by its power series. The associated continuous argument of $γ - p$ along $γ$ is $θ : = Im λ : [a, b] \to R$ (Real and imaginary parts, complex conjugation, and modulus).

Let $V \subseteq C$ be open with $p \notin V$ . A holomorphic logarithm branch of $z - p$ on $V$ is a holomorphic function $L : V \to C$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions) with

$\exp (L (z)) = z - p for every z \in V .$

Remarks

These are two different objects and only the first is unconditional. A continuous logarithm along $γ$ is a function of the parameter $t$ ; it exists for every complex contour missing $p$ , and prescribing the single value $λ (a)$ among the complex numbers whose exponential is $γ (a) - p$ determines it, both by Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$ ↗. A holomorphic logarithm branch is a function on a plane set, and for a general open $V$ missing $p$ there need be none.

Along a contour, whose parameter interval is connected, two continuous logarithms differ by one additive constant in $2 π i Z$ . On a general open set $V$ , two holomorphic branches differ by a locally constant $2 π i Z$ -valued function, hence by one such constant on each connected component; a single global constant is forced only when $V$ is connected. This follows from $\ker (\exp) = 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ). For $p = 0$ and $V$ the slit plane, Complex logarithms, the principal logarithm, and principal and multivalued complex powers names the principal logarithm; its holomorphy on that domain is proved later on this page.

A continuous argument $θ$ carries no normalisation of its own: adding $2 π k$ to $θ$ for a fixed integer $k$ replaces $λ$ by $λ + 2 π i k$ , which is again a continuous logarithm. What is unambiguous is the increment $θ (b) - θ (a)$ , since the two choices differ by the same constant at both endpoints.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$

Statement

Let $γ : [a, b] \to C$ be a complex contour and let $p \in C$ with $p \notin γ^{*}$ . Then:

there is a continuous logarithm $λ$ of $γ - p$ along $γ$ (Continuous logarithms and continuous arguments along a contour);
if $λ_{1}$ and $λ_{2}$ are two of them, then $λ_{1} - λ_{2}$ is a constant function with value in $2 π i Z$ ;
for each $v \in C$ with $\exp v = γ (a) - p$ there is exactly one continuous logarithm $λ$ of $γ - p$ along $γ$ with $λ (a) = v$ .

In particular the increment $λ (b) - λ (a)$ , and the increment $θ (b) - θ (a)$ of the associated continuous argument, are the same for every choice of $λ$ . No differentiability of $γ$ is used.

Facts & Assumptions

Given: A complex contour $γ : [a, b] \to C$ and a point $p \notin γ^{*}$ .

[L1]

A continuous logarithm of $γ - p$ along $γ$ is a continuous $λ : [a, b] \to C$ with $\exp (λ (t)) = γ (t) - p$ for every $t$ ; a holomorphic logarithm branch of $z - p$ on an open $V$ missing $p$ is a holomorphic $L$ on $V$ with $\exp (L (z)) = z - p$ (Continuous logarithms and continuous arguments along a contour).

[L2]

For a complex contour $γ$ and $p \notin γ^{*}$ , the distance $d = \inf {∣ w - p ∣ : w \in γ^{*}}$ is positive and there is $δ > 0$ such that every partition $a = t_{0} < \dots < t_{r} = b$ of mesh below $δ$ has $γ ([t_{i}, t_{i + 1}]) \subseteq D (γ (t_{i}), d)$ and $p \notin D (γ (t_{i}), d)$ for every $i < r$ ; at least one such partition exists (A contour missing a point subdivides into arcs lying in discs that miss it).

[L3]

If $D (c, ρ)$ is an open disc with $ρ > 0$ and $p \notin D (c, ρ)$ , there is a holomorphic $L$ on $D (c, ρ)$ with $\exp (L (z)) = z - p$ there (A disc missing $p$ carries a holomorphic logarithm of $z - p$ ).

[L4]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L5]

The complex exponential maps $C$ onto $C ∖ {0}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

[L6]

$\exp (z + w) = \exp z \exp w$ for all $z, w \in C$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[L7]

The continuous image of a connected subset is a connected subset (A continuous image of a connected space is connected, and connectedness is a topological property).

[L8]

A subset $E \subseteq R$ is connected exactly when it is order-convex (The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ "); a closed bounded interval is order-convex (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L9]

If a property holds at $0$ and passes from $n$ to $n + 1$ , it holds for every natural number (The principle of mathematical induction).

[L10]

A composite of continuous maps is continuous, and a function whose restrictions to the members of a finite closed cover are continuous is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L11]

For $z = a + b i$ with $a, b$ real, $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

[L12]

A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L13]

Proof

technique · direct

1.1givenL4L5

Since $p \notin γ^{*}$ , the number $γ (a) - p$ is nonzero, so [L5] supplies $v \in C$ with $\exp v = γ (a) - p$ ; more generally, for every such $v$ the set of complex numbers with that exponential is $v + 2 π i Z$ by [L4].

1.2L1L4L7L8L10L11L13

If $λ_{1}, λ_{2}$ are continuous logarithms of $γ - p$ along $γ$ , then $\exp (λ_{1} (t)) = \exp (λ_{2} (t))$ for every $t$ , so $λ_{1} (t) - λ_{2} (t) \in 2 π i Z$ by [L4]; the real-valued function $g = Im (λ_{1} - λ_{2}) / (2 π)$ is continuous by [L10] and [L11] and takes values in $Z$ , so by [L7] and [L8] its image is an order-convex subset of $R$ inside $Z$ , which by [L13] can only be a single point. Hence $λ_{1} - λ_{2}$ is a constant in $2 π i Z$ , and it is $0$ when $λ_{1} (a) = λ_{2} (a)$ .

1.3givenL2

Assume $a < b$ . By [L2] there are $d > 0$ and a partition $a = t_{0} < t_{1} < \dots < t_{r} = b$ with $γ ([t_{i}, t_{i + 1}]) \subseteq D_{i} : = D (γ (t_{i}), d)$ and $p \notin D_{i}$ for every $i < r$ .

2.1step 1.3L3L12

By [L3] each $D_{i}$ carries a holomorphic $L_{i}$ with $\exp (L_{i} (z)) = z - p$ for $z \in D_{i}$ , and $L_{i}$ is continuous on $D_{i}$ by [L12].

2.2step 1.1step 1.3

Define $c_{0} : = v$ and, for each $i < r$ , define $c_{i + 1} : = c_{i} + L_{i} (γ (t_{i + 1})) - L_{i} (γ (t_{i}))$ ; this determines the finite list $c_{0}, \dots, c_{r}$ . Now define $λ$ on $[t_{i}, t_{i + 1}]$ by $λ (t) = c_{i} + L_{i} (γ (t)) - L_{i} (γ (t_{i}))$ . The two formulas available at a shared point $t_{i}$ with $0 < i < r$ agree, the $i$ th giving $c_{i}$ and the $(i - 1)$ st giving $c_{i - 1} + L_{i - 1} (γ (t_{i})) - L_{i - 1} (γ (t_{i - 1})) = c_{i}$ , so $λ : [a, b] \to C$ is a well-defined function with $λ (a) = v$ and $λ (t_{i}) = c_{i}$ for every $i \leq r$ .

3.1step 2.1step 2.2L10

Each restriction $λ ∣_{[t_{i}, t_{i + 1}]}$ is continuous, being a constant plus the composite of $γ$ with $L_{i}$ of step 2.1; the intervals $[t_{i}, t_{i + 1}]$ form a finite closed cover of $[a, b]$ , so $λ$ is continuous by [L10].

3.2step 1.1step 2.1step 2.2L6L9

For every $i < r$ and $t \in [t_{i}, t_{i + 1}]$ , [L6] gives $\exp (λ (t)) = \exp (c_{i}) \exp (L_{i} (γ (t))) \exp (L_{i} (γ (t_{i})))^{- 1}$ , and $\exp (L_{i} (γ (t))) = γ (t) - p$ by step 2.1, so $\exp (c_{i}) = γ (t_{i}) - p$ forces $\exp (λ (t)) = γ (t) - p$ and, at $t = t_{i + 1}$ , $\exp (c_{i + 1}) = γ (t_{i + 1}) - p$ . Since $\exp (c_{0}) = \exp (v) = γ (a) - p$ , an induction on $i$ ([L9]) gives $\exp (λ (t)) = γ (t) - p$ for every $t \in [a, b]$ .

4.1step 1.2step 3.1step 3.2L1L11∎

Steps 3.1 and 3.2 make $λ$ a continuous logarithm of $γ - p$ along $γ$ with $λ (a) = v$ , which proves claims 1 and 3 when $a < b$ ; when $a = b$ the constant function with value $v$ does the same, since its only value satisfies $\exp v = γ (a) - p$ . Claim 2 is step 1.2, which also gives the uniqueness in claim 3, and it makes $λ (b) - λ (a)$ and its imaginary part independent of the choice by [L1] and [L11].

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm

Statement

Let $γ : [a, b] \to C$ be a complex contour, let $p \in C$ with $p \notin γ^{*}$ , and let $λ$ be a continuous logarithm of $γ - p$ along $γ$ (Continuous logarithms and continuous arguments along a contour). Then

$\int_{γ} \frac{d z}{z - p} = λ (b) - λ (a) .$

The contour need not be closed, and the right-hand side is the same for every continuous logarithm of $γ - p$ along $γ$ .

Facts & Assumptions

Given: A complex contour $γ : [a, b] \to C$ , a point $p \notin γ^{*}$ , and a continuous logarithm $λ$ of $γ - p$ along $γ$ .

[L1]

A continuous logarithm of $γ - p$ along $γ$ is a continuous $λ : [a, b] \to C$ with $\exp (λ (t)) = γ (t) - p$ for every $t$ (Continuous logarithms and continuous arguments along a contour).

[L2]

For a complex contour $γ$ and $p \notin γ^{*}$ there is a continuous logarithm of $γ - p$ along $γ$ , and any two of them differ by a constant lying in $2 π i Z$ (Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$ ).

[L3]

For a complex contour $γ$ and $p \notin γ^{*}$ , the distance $d = \inf {∣ w - p ∣ : w \in γ^{*}}$ is positive and some partition $a = t_{0} < \dots < t_{r} = b$ satisfies $γ ([t_{i}, t_{i + 1}]) \subseteq D (γ (t_{i}), d)$ and $p \notin D (γ (t_{i}), d)$ for every $i < r$ (A contour missing a point subdivides into arcs lying in discs that miss it).

[L4]

If $D (c, ρ)$ is an open disc with $ρ > 0$ and $p \notin D (c, ρ)$ , there is a holomorphic $L$ on $D (c, ρ)$ with $\exp (L (z)) = z - p$ there, and every such $L$ satisfies $L^{'} (z) = 1 / (z - p)$ (A disc missing $p$ carries a holomorphic logarithm of $z - p$ ).

[L5]

If $F$ is a primitive of a continuous $f$ on an open set containing the trace of a rectifiable contour $γ : [a, b] \to C$ and $F^{'} = f$ is continuous, then $\int_{γ} f (z) d z = F (γ (b)) - F (γ (a))$ (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path, A primitive of a complex function on an open set).

[L6]

If $ϕ : [c, d] \to [a, b]$ is a strictly increasing continuous bijection and $f$ is continuous on the trace of the rectifiable $γ$ , then $\int_{γ \circ ϕ} f d z = \int_{γ} f d z$ (Complex and absolute line integrals are invariant under increasing continuous reparametrization).

[L7]

For composable rectifiable contours $α, β$ , $\int_{α * β} f d z = \int_{α} f d z + \int_{β} f d z$ (Complex line integrals change sign under reversal and add under concatenation); concatenation of $α, β : [0, 1] \to C$ with $α (1) = β (0)$ is $(α * β) (s) = α (2 s)$ for $s \leq \frac{1}{2}$ and $β (2 s - 1)$ for $s \geq \frac{1}{2}$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L8]

For a rectifiable $γ$ and $f$ continuous on its trace, $\int_{γ} f d z$ exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L9]

Arc length is additive across a split of the parameter interval, and $γ$ is rectifiable exactly when both restrictions are (Arc length is additive across every subdivision point and decreases under restriction).

[L10]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L11]

The continuous image of a connected subset is connected (A continuous image of a connected space is connected, and connectedness is a topological property), and a connected subset of $R$ is order-convex (The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ ").

[L12]

If a property holds at $0$ and passes from $n$ to $n + 1$ , it holds for every natural number (The principle of mathematical induction).

[L14]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous), and a function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L15]

For $z = a + b i$ with $a, b$ real, $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

[L16]

[L17]

Nonvanishing quotients of functions complex differentiable at a point are complex differentiable there (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Proof

technique · direct

1.1givenL8L14L17

The function $z \mapsto 1 / (z - p)$ is defined and continuous on $γ^{*}$ by [L14] and [L17], since $p \notin γ^{*}$ , so the integral $\int_{γ} d z / (z - p)$ exists by [L8].

1.2L2

By [L2] any two continuous logarithms of $γ - p$ along $γ$ differ by a constant, so the increment $λ (b) - λ (a)$ is the same for all of them.

1.3givenL3L4L9

Assume $a < b$ . By [L3] fix $d > 0$ and a partition $a = t_{0} < \dots < t_{r} = b$ with $γ ([t_{i}, t_{i + 1}]) \subseteq D_{i} : = D (γ (t_{i}), d)$ and $p \notin D_{i}$ for $i < r$ , and by [L4] fix a holomorphic $L_{i}$ on $D_{i}$ with $\exp (L_{i} (z)) = z - p$ and $L_{i}^{'} (z) = 1 / (z - p)$ there. By [L9] each restriction $γ_{i} : = γ ∣_{[t_{i}, t_{i + 1}]}$ is rectifiable.

2.1givenstep 1.3L1L10L11L14L15L16

Fix $i < r$ . For $t \in [t_{i}, t_{i + 1}]$ both $\exp (λ (t))$ and $\exp (L_{i} (γ (t)))$ equal $γ (t) - p$ , so $λ (t) - L_{i} (γ (t)) \in 2 π i Z$ by [L10]; that difference is continuous by [L14], its scaled imaginary part is a continuous integer-valued real function by [L15], and [L11] with [L16] forces it to be constant on the interval. Hence $λ (t_{i + 1}) - λ (t_{i}) = L_{i} (γ (t_{i + 1})) - L_{i} (γ (t_{i}))$ .

2.2step 1.3L4L5L14L17

Fix $i < r$ . The trace of $γ_{i}$ lies in the open disc $D_{i}$ , on which $L_{i}$ is a primitive of the continuous function $1 / (z - p)$ , so [L5] gives $\int_{γ_{i}} d z / (z - p) = L_{i} (γ (t_{i + 1})) - L_{i} (γ (t_{i}))$ .

2.3step 1.1step 1.3L6L7L12

For $a \leq u < v < w \leq b$ the increasing affine reparametrisations $α (s) = γ (u + s (v - u))$ and $β (s) = γ (v + s (w - v))$ of $[0, 1]$ satisfy $α (1) = β (0)$ and $α * β = γ ∣_{[u, w]} \circ ϕ$ for the strictly increasing continuous bijection $ϕ : [0, 1] \to [u, w]$ that is affine on $[0, \frac{1}{2}]$ and on $[\frac{1}{2}, 1]$ with $ϕ (\frac{1}{2}) = v$ , so [L6] and [L7] split the integral at $v$ ; applying this at $t_{1}$ , then to $γ ∣_{[t_{1}, b]}$ at $t_{2}$ , and so on, an induction on the number of partition points ([L12]) gives $\int_{γ} d z / (z - p) = \sum_{i < r} \int_{γ_{i}} d z / (z - p)$ .

3.1step 2.1step 2.2step 2.3algebra

Substituting step 2.2 into step 2.3 and then step 2.1, the integral equals $\sum_{i < r} (λ (t_{i + 1}) - λ (t_{i}))$ . Expanding this finite sum, every intermediate value $λ (t_{i})$ with $0 < i < r$ appears once with sign $+$ and once with sign $-$ , so the sum telescopes to $λ (t_{r}) - λ (t_{0}) = λ (b) - λ (a)$ .

4.1step 1.2L4L5∎

If instead $a = b$ , choose $ρ > 0$ with $p \notin D (γ (a), ρ)$ ; [L4] gives a holomorphic $L$ on that disc with $L^{'} (z) = 1 / (z - p)$ . The trace of the constant contour $γ$ lies in that disc, so [L5] gives $\int_{γ} d z / (z - p) = L (γ (a)) - L (γ (a)) = 0$ , while $λ (b) - λ (a) = 0$ . Thus the identity also holds when $a = b$ ; and by step 1.2 the value asserted is independent of which continuous logarithm is used.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The winding number of a closed contour about a point off its trace

Definition

Let $γ : [a, b] \to C$ be a closed complex contour, that is a rectifiable path with $γ (a) = γ (b)$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation), with trace $γ^{*}$ , and let $p \in C$ with $p \notin γ^{*}$ . The winding number, or index, of $γ$ about $p$ is

$n (γ, p) : = \frac{1}{2 π i} \int_{γ} \frac{d z}{z - p},$

the complex line integral of The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral.

The integral exists: $z \mapsto 1 / (z - p)$ is complex differentiable, hence continuous, on $C ∖ {p}$ by Linearity, product, reciprocal, and quotient rules for complex derivatives and Complex differentiability at a point implies continuity there, the trace $γ^{*}$ is contained in that set, and $γ$ is rectifiable, so Continuous integrands have complex and absolute line integrals along every rectifiable path applies.

Remarks

The index is attached to the parametrised contour and not to its trace. Two closed contours with the same trace can have different indices about the same point, because the parametrisation records how many times, and in which direction, the trace is traversed; the definition above reads $γ$ as a map and the integral depends on that map.

The point $p$ is required to lie off the trace. On the trace the integrand $1 / (z - p)$ is undefined at $z = p$ , so no value $n (γ, p)$ is defined there and none is asserted anywhere below.

No connectedness is assumed of the set $C ∖ γ^{*}$ where the index lives; when a complex domain (A complex domain is a nonempty connected open subset of $C$ ) is wanted it is said so explicitly.

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The winding number of a closed contour is an integer

Statement

Let $γ : [a, b] \to C$ be a closed complex contour and let $p \in C$ with $p \notin γ^{*}$ . Then

$n (γ, p) = \frac{1}{2 π i} \int_{γ} \frac{d z}{z - p} \in Z .$

No differentiability of $γ$ is used: the contour is only assumed rectifiable.

Facts & Assumptions

Given: A closed complex contour $γ : [a, b] \to C$ and a point $p \notin γ^{*}$ .

[L1]

For a closed complex contour $γ$ and $p \notin γ^{*}$ , $n (γ, p) = (2 π i)^{- 1} \int_{γ} d z / (z - p)$ (The winding number of a closed contour about a point off its trace).

[L2]

[L3]

For a complex contour $γ$ and $p \notin γ^{*}$ there is a continuous logarithm of $γ - p$ along $γ$ (Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$ ), namely a continuous $λ : [a, b] \to C$ with $\exp (λ (t)) = γ (t) - p$ for every $t$ (Continuous logarithms and continuous arguments along a contour).

[L4]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L5]

A complex contour is closed when $γ (a) = γ (b)$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

[L6]

$Z$ is the ring of integers (The integers as equivalence classes of pairs of naturals).

Proof

technique · direct

1.1givenL1L2L3

By [L3] fix a continuous logarithm $λ$ of $γ - p$ along $γ$ ; then [L1] and [L2] give $2 π i n (γ, p) = λ (b) - λ (a)$ .

1.2givenL3L5

Since $γ$ is closed, $γ (b) = γ (a)$ by [L5], so $\exp (λ (b)) = γ (b) - p = γ (a) - p = \exp (λ (a))$ .

2.1step 1.1step 1.2L4

By [L4] the equality of exponentials in step 1.2 gives $λ (b) - λ (a) \in 2 π i Z$ , so step 1.1 makes $2 π i n (γ, p)$ an element of $2 π i Z$ .

3.1step 2.1L2L3L6∎

Dividing by $2 π i$ in step 2.1 puts $n (γ, p)$ in $Z$ by [L6]. The argument used only the rectifiability of $γ$ , through [L2] and [L3], and never a derivative of $γ$ .

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The winding number is the increment of a continuous argument divided by $2 π$

Statement

Let $γ : [a, b] \to C$ be a closed complex contour, let $p \notin γ^{*}$ , let $λ$ be a continuous logarithm of $γ - p$ along $γ$ and let $θ = Im λ$ be the associated continuous argument (Continuous logarithms and continuous arguments along a contour). Then

$Re λ (t) = \log ∣ γ (t) - p ∣ (a \leq t \leq b), n (γ, p) = \frac{θ (b) - θ (a)}{2 π} .$

In particular $θ (b) - θ (a)$ is an integer multiple of $2 π$ , and it is the same for every continuous argument of $γ - p$ along $γ$ .

Facts & Assumptions

Given: A closed complex contour $γ : [a, b] \to C$ , a point $p \notin γ^{*}$ , a continuous logarithm $λ$ of $γ - p$ along $γ$ , and $θ = Im λ$ .

[L1]

For a closed complex contour and $p$ off its trace, $n (γ, p) \in Z$ (The winding number of a closed contour is an integer), where $n (γ, p) = (2 π i)^{- 1} \int_{γ} d z / (z - p)$ (The winding number of a closed contour about a point off its trace).

[L2]

A continuous logarithm of $γ - p$ along $γ$ is a continuous $λ$ with $\exp (λ (t)) = γ (t) - p$ for every $t$ , its continuous argument is $θ = Im λ$ , and any two continuous logarithms differ by a constant in $2 π i Z$ (Continuous logarithms and continuous arguments along a contour, Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$ ).

[L3]

For real $x, y$ , $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L4]

For a complex contour $γ$ , $p \notin γ^{*}$ and a continuous logarithm $λ$ of $γ - p$ along $γ$ , $\int_{γ} d z / (z - p) = λ (b) - λ (a)$ (The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm).

[L5]

For $x > 0$ , $\log x$ is the unique real $y$ with $\exp y = x$ (The natural logarithm as the inverse of the exponential function).

[L6]

For $z = a + b i$ with $a, b$ real, $Re z = a$ and $Im z = b$ (Real and imaginary parts, complex conjugation, and modulus).

[L7]

A complex contour is closed when $γ (a) = γ (b)$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

[L8]

The integers form a commutative ring (The integers form a commutative ring).

Proof

technique · direct

1.1givenL2L3L5L6

Writing $λ (t) = Re λ (t) + i θ (t)$ as in [L6], the identity $\exp (λ (t)) = γ (t) - p$ of [L2] and the modulus formula [L3] give $∣ γ (t) - p ∣ = e^{Re λ (t)}$ , so $Re λ (t) = \log ∣ γ (t) - p ∣$ by [L5].

1.2givenL7

Since $γ$ is closed, $γ (b) = γ (a)$ by [L7], so $∣ γ (b) - p ∣ = ∣ γ (a) - p ∣$ .

2.1step 1.1step 1.2L6

Steps 1.1 and 1.2 give $Re λ (b) = Re λ (a)$ , hence $λ (b) - λ (a) = i (θ (b) - θ (a))$ by [L6].

3.1step 2.1L1L4

By [L1] and [L4], $2 π i n (γ, p) = λ (b) - λ (a)$ , which step 2.1 rewrites as $i (θ (b) - θ (a))$ ; dividing by $2 π i$ gives $n (γ, p) = (θ (b) - θ (a)) / (2 π)$ .

4.1step 3.1L1L2L8∎

Since $n (γ, p)$ is an integer by [L1] and [L8], step 3.1 makes $θ (b) - θ (a) = 2 π n (γ, p)$ an integer multiple of $2 π$ ; and replacing $λ$ by another continuous logarithm changes it by a constant of $2 π i Z$ by [L2], which cancels in the increment, so the value is the same for every continuous argument.

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The winding number is constant on each connected component of the complement of the trace

Statement

Let $γ$ be a closed complex contour with trace $γ^{*}$ and length $L (γ)$ . Then $C ∖ γ^{*}$ is open, and the index function $p \mapsto n (γ, p)$ satisfies the quantitative estimate

$∣ n (γ, p) - n (γ, p_{0}) ∣ \leq \frac{L (γ) ∣ p - p_{0} ∣}{π d^{2}} whenever p_{0} \notin γ^{*}, d = \inf_{w \in γ^{*}} ∣ w - p_{0} ∣, ∣ p - p_{0} ∣ < \frac{d}{2} .$

In particular $n (γ, \cdot)$ is continuous on $C ∖ γ^{*}$ ; it is constant on every connected component of that set; and since those components are open, it is locally constant.

Facts & Assumptions

Given: A closed complex contour $γ : [a, b] \to C$ ; the plane carries the Euclidean metric of $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

For a closed complex contour $γ$ and $p \notin γ^{*}$ , $n (γ, p) = (2 π i)^{- 1} \int_{γ} d z / (z - p)$ (The winding number of a closed contour about a point off its trace).

[L2]

The winding number of a closed complex contour about a point off its trace is an integer (The winding number of a closed contour is an integer).

[L3]

For a complex contour $γ$ and $p_{0} \notin γ^{*}$ , the distance $d = \inf {∣ w - p_{0} ∣ : w \in γ^{*}}$ exists and is positive (A contour missing a point subdivides into arcs lying in discs that miss it).

[L4]

[L5]

[L6]

The connected component $C (x)$ is the union of all connected subsets containing $x$ , hence the largest connected subset containing $x$ (Connected components, quasicomponents, and totally disconnected spaces).

[L7]

Every connected component of an open subset $U \subseteq R^{n}$ is open in $R^{n}$ and polygonally connected (Every connected component of an open subset of $R^{n}$ is open and polygonally connected).

[L8]

[L9]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded); the continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); and a closed bounded interval is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L10]

A set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and $B (x, r) = {y : d (x, y) < r}$ (Open ball, closed ball and sphere in a metric space).

[L11]

[L12]

Proof

technique · direct

1.1givenL9L10

The trace $γ^{*}$ is the continuous image of a compact interval, hence compact and closed by [L9], so its complement $C ∖ γ^{*}$ is open by [L10].

1.2givenL3L11

Fix $p_{0} \notin γ^{*}$ and put $d = \inf {∣ w - p_{0} ∣ : w \in γ^{*}}$ , which is positive by [L3]; let $p$ satisfy $∣ p - p_{0} ∣ < d / 2$ . For $z \in γ^{*}$ one has $∣ z - p_{0} ∣ \geq d$ and, by [L11], $∣ z - p ∣ \geq ∣ z - p_{0} ∣ - ∣ p - p_{0} ∣ > d - \frac{d}{2} = \frac{d}{2} > 0$ , so $p \notin γ^{*}$ as well.

2.1step 1.2L11algebra

For $z \in γ^{*}$ , elementary algebra gives $\frac{1}{z - p} - \frac{1}{z - p_{0}} = \frac{p - p_{0}}{(z - p) (z - p_{0})}$ , whose modulus is at most $∣ p - p_{0} ∣ / (\frac{d}{2} \cdot d) = 2 ∣ p - p_{0} ∣ / d^{2}$ by step 1.2 and [L11].

3.1step 2.1L1L4L5

By [L1] and [L5] the difference $n (γ, p) - n (γ, p_{0})$ is $(2 π i)^{- 1} \int_{γ} (\frac{1}{z - p} - \frac{1}{z - p_{0}}) d z$ , and [L4] with the bound of step 2.1 makes its modulus at most $(2 π)^{- 1} \cdot 2 ∣ p - p_{0} ∣ L (γ) / d^{2} = L (γ) ∣ p - p_{0} ∣ / (π d^{2})$ .

4.1step 3.1L10

Step 3.1 shows $n (γ, \cdot)$ is continuous at every $p_{0} \notin γ^{*}$ , since the bound tends to $0$ with $∣ p - p_{0} ∣$ .

5.1step 1.1step 4.1L2L6L7L8L12∎

Let $C$ be a connected component of $C ∖ γ^{*}$ . By [L2] the function $n (γ, \cdot)$ is integer-valued, and by step 4.1 it is continuous, so by [L6] and [L8] its image on $C$ is an order-convex subset of $R$ contained in $Z$ ; by [L12] such a set has at most one element, so $n (γ, \cdot)$ is constant on $C$ . By [L7] applied to the open set of step 1.1, $C$ is open, so the index is locally constant on $C ∖ γ^{*}$ .

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The exterior of a closed disc in the plane is path-connected

Statement

Let $c \in C$ . Then:

for every real $R \geq 0$ , the open exterior $E_{>} : = {z \in C : ∣ z - c ∣ > R}$ is path-connected, and therefore a connected subset of $C$ ; taking $R = 0$ , the punctured plane $C ∖ {c}$ is path-connected;
for every real $R > 0$ , the closed exterior $E_{\geq} : = {z \in C : ∣ z - c ∣ \geq R}$ is path-connected, and therefore a connected subset of $C$ .

Facts & Assumptions

Given: A point $c \in C$ and a real $R$ , with $R \geq 0$ in clause 1 and $R > 0$ in clause 2; the plane is read as $R^{2}$ with its Euclidean metric through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves. Write $E$ for whichever of the two sets is under discussion.

[L1]

For $n \geq 2$ the unit sphere $S^{n - 1} \subseteq R^{n}$ is path-connected and connected (For $n \geq 2$ , the sphere $S^{n - 1}$ is path-connected and connected).

[L2]

For $n \geq 1$ the map $ρ (x) = x / ∥ x ∥_{2}$ from $R^{n} ∖ {0}$ to $S^{n - 1}$ is continuous (Radial normalisation $x \mapsto x / ∥ x ∥_{2}$ is continuous on $R^{n} ∖ {0}$ ).

[L3]

For $n \geq 1$ , $S^{n - 1} = S_{2} (0, 1) = {x \in R^{n} : ∥ x ∥_{2} = 1}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[L4]

A subset is path-connected when any two of its points are joined by a continuous map from $[0, 1]$ whose image lies in it (Paths, path-connected spaces and path components).

[L5]

A path-connected subset of a topological space is a connected subset (Every path-connected space is connected, and every path component lies inside a component).

[L6]

[L7]

$B (x, r) = {y : d (x, y) < r}$ (Open ball, closed ball and sphere in a metric space).

[L8]

Proof

technique · direct

1.1givenL2L3L7

Let $z, w \in E$ and put $ρ = \max (∣ z - c ∣, ∣ w - c ∣)$ . In clause 1 this gives $ρ \geq ∣ z - c ∣ > R \geq 0$ and $ρ \geq ∣ w - c ∣ > R$ , and in clause 2 it gives $ρ \geq ∣ z - c ∣ \geq R > 0$ and $ρ \geq ∣ w - c ∣ \geq R$ ; in both cases $ρ > 0$ and $z \neq c$ , $w \neq c$ , so $u = (z - c) / ∣ z - c ∣$ and $v = (w - c) / ∣ w - c ∣$ lie on the unit circle $S^{1}$ by [L2] and [L3].

1.2L1L4

By [L1] with $n = 2$ there is a continuous $σ : [0, 1] \to S^{1}$ with $σ (0) = u$ and $σ (1) = v$ .

2.1step 1.1L6L7L8

The map $μ (s) = c + ((1 - s) ∣ z - c ∣ + s ρ) u$ is continuous on $[0, 1]$ by [L6], joins $z$ to $c + ρ u$ , and satisfies $∣ μ (s) - c ∣ = (1 - s) ∣ z - c ∣ + s ρ$ by [L8] and $∣ u ∣ = 1$ ; that value lies between $∣ z - c ∣$ and $ρ$ , so it exceeds $R$ in clause 1 and is at least $R$ in clause 2, and $μ$ has image in $E$ . The same formula with $w$ and $v$ gives a continuous $ν : [0, 1] \to E$ joining $w$ to $c + ρ v$ .

2.2step 1.1step 1.2L6L7L8

The map $s \mapsto c + ρ σ (s)$ is continuous on $[0, 1]$ by [L6], joins $c + ρ u$ to $c + ρ v$ , and has $∣ c + ρ σ (s) - c ∣ = ρ$ by [L8], which exceeds $R$ in clause 1 and is at least $R$ in clause 2, so its image lies in $E$ .

3.1step 2.1step 2.2L4L5L6∎

Concatenating $μ$ , the path of step 2.2 and the reversal of $ν$ , each on a closed subinterval of $[0, 1]$ and agreeing at the two shared endpoints, gives by [L6] a continuous map $[0, 1] \to E$ from $z$ to $w$ . Since $z, w \in E$ were arbitrary, $E$ is path-connected by [L4], hence a connected subset of $C$ by [L5]; the argument was run for both clauses at once, and at $R = 0$ clause 1 reads $C ∖ {c}$ .

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The complement of a compact plane set has exactly one unbounded connected component

Statement

Let $K \subseteq C$ be compact. Then $C ∖ K$ has exactly one unbounded connected component $U_{\infty}$ , and every other component is bounded. Moreover, whenever $R > 0$ satisfies $K \subseteq {z : ∣ z ∣ \leq R}$ , the exterior ${z : ∣ z ∣ > R}$ is contained in $U_{\infty}$ .

The empty set is covered: $C ∖ \emptyset = C$ has the single component $C$ , which is unbounded.

Facts & Assumptions

Given: A compact set $K \subseteq C$ ; the plane is read as $R^{2}$ with its Euclidean metric through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

For $c \in C$ and $R > 0$ , the set ${z : ∣ z - c ∣ > R}$ is path-connected and a connected subset of $C$ (The exterior of a closed disc in the plane is path-connected).

[L2]

The connected component $C (x)$ is the union of all connected subsets containing $x$ (Connected components, quasicomponents, and totally disconnected spaces).

[L3]

$C (x)$ contains every connected subset containing $x$ ; distinct components are disjoint; every point lies in its own component, and the components cover the space (The components of a space are its maximal connected subsets, they partition it, and each of them is closed).

[L4]

A union of connected subsets with a common point is connected (A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member).

[L5]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

[L6]

A subset $A$ of a metric space is bounded when $A = \emptyset$ or $A \subseteq B (x_{0}, r)$ for some $x_{0}$ and some real $r > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

[L7]

A set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L8]

For $n \geq 1$ , $R^{n}$ is polygonally connected and connected ( $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected).

[L9]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Proof

technique · direct

1.1givenL5L6L7

By [L5] the set $K$ is closed and bounded, so by [L6] there are $x_{0}$ and $s > 0$ with $K \subseteq B (x_{0}, s)$ , and then $K \subseteq {z : ∣ z ∣ \leq R}$ for $R = ∣ x_{0} ∣ + s > 0$ ; if $K = \emptyset$ any $R > 0$ serves. By [L7] the complement $C ∖ K$ is open.

1.2givenL1L6L9

Fix any $R > 0$ with $K \subseteq {z : ∣ z ∣ \leq R}$ and put $E_{R} = {z : ∣ z ∣ > R}$ . Then $E_{R} \subseteq C ∖ K$ , and $E_{R}$ is a connected subset of $C$ by [L1]; it is nonempty, since $2 R \in E_{R}$ , and unbounded, since for every real $r > 0$ and every $x_{0}$ the number $R + ∣ x_{0} ∣ + r$ has modulus exceeding $R$ and lies outside $B (x_{0}, r)$ , so no ball of [L6] contains $E_{R}$ .

2.1step 1.1step 1.2L2L3L6

Let $z_{0} \in E_{R}$ and let $U_{\infty} : = C (z_{0})$ be its component in $C ∖ K$ . The set $E_{R}$ is a connected subset of $C ∖ K$ containing $z_{0}$ , so $E_{R} \subseteq U_{\infty}$ by [L2] and [L3]; hence $U_{\infty}$ is unbounded by step 1.2 and [L6]. This holds for every admissible $R$ , which is the final clause of the statement.

3.1step 2.1L3L6

Let $C$ be a component of $C ∖ K$ with $C \neq U_{\infty}$ . By [L3] the two are disjoint, so $C \cap E_{R} = \emptyset$ by step 2.1, that is $C \subseteq {z : ∣ z ∣ \leq R} \subseteq B (0, R + 1)$ ; hence $C$ is bounded by [L6].

4.1step 2.1step 3.1L3L4L8∎

Steps 2.1 and 3.1 give exactly one unbounded component, namely $U_{\infty}$ , with every other component bounded. When $K = \emptyset$ the complement is $C$ , which is connected by [L8], so by [L3] and [L4] it is its own single component and that component is $U_{\infty}$ .

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The winding number vanishes on the unbounded component of the complement of the trace

Statement

Let $γ$ be a closed complex contour with trace $γ^{*}$ and length $L (γ)$ . Then $C ∖ γ^{*}$ has exactly one unbounded connected component $U_{\infty}$ , and

$n (γ, p) = 0 for every p \in U_{\infty} .$

More precisely, if $R > 0$ satisfies $γ^{*} \subseteq {z : ∣ z ∣ \leq R}$ and $∣ p ∣ > R + L (γ) / (2 π)$ , then $p \in U_{\infty}$ and $n (γ, p) = 0$ .

Facts & Assumptions

Given: A closed complex contour $γ : [a, b] \to C$ .

[L1]

For a compact $K \subseteq C$ , the complement $C ∖ K$ has exactly one unbounded connected component $U_{\infty}$ , every other component is bounded, and ${z : ∣ z ∣ > R} \subseteq U_{\infty}$ whenever $R > 0$ satisfies $K \subseteq {z : ∣ z ∣ \leq R}$ (The complement of a compact plane set has exactly one unbounded connected component).

[L2]

The index $n (γ, \cdot)$ is constant on every connected component of $C ∖ γ^{*}$ (The winding number is constant on each connected component of the complement of the trace).

[L3]

The winding number of a closed complex contour about a point off its trace is an integer (The winding number of a closed contour is an integer).

[L4]

[L5]

$n (γ, p) = (2 π i)^{- 1} \int_{γ} d z / (z - p)$ (The winding number of a closed contour about a point off its trace).

[L6]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded); the continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); a closed bounded interval is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L7]

The connected component $C (x)$ is the union of all connected subsets containing $x$ (Connected components, quasicomponents, and totally disconnected spaces) and contains every connected subset containing $x$ (The components of a space are its maximal connected subsets, they partition it, and each of them is closed).

[L8]

A subset of a metric space is bounded when it is empty or contained in some ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L9]

$∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for complex $z, w$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L10]

The integers form an ordered commutative ring and their canonical image in $R$ is discrete; in particular the only integer of modulus below $1$ is $0$ (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

Proof

technique · direct

1.1givenL1L6L8

The trace $γ^{*}$ is the continuous image of a compact interval, hence compact by [L6], and bounded by [L6], so there is $R > 0$ with $γ^{*} \subseteq {z : ∣ z ∣ \leq R}$ . By [L1] the set $C ∖ γ^{*}$ has exactly one unbounded component $U_{\infty}$ , and ${z : ∣ z ∣ > R} \subseteq U_{\infty}$ .

2.1step 1.1L4L5L9

Let $∣ p ∣ > R$ . For $w \in γ^{*}$ one has $∣ w ∣ \leq R$ , so $∣ w - p ∣ \geq ∣ p ∣ - ∣ w ∣ \geq ∣ p ∣ - R > 0$ by [L9]; hence $p \notin γ^{*}$ and $∣ 1 / (z - p) ∣ \leq 1 / (∣ p ∣ - R)$ on the trace. By [L4] and [L5], $∣ n (γ, p) ∣ \leq L (γ) / (2 π (∣ p ∣ - R))$ .

3.1step 1.1step 2.1L3L10

If in addition $∣ p ∣ > R + L (γ) / (2 π)$ then $∣ p ∣ - R > L (γ) / (2 π)$ , so step 2.1 gives $∣ n (γ, p) ∣ < 1$ ; since $n (γ, p)$ is an integer by [L3], it is $0$ by [L10]. Such $p$ exist, for instance $p = R + L (γ) / (2 π) + 1$ , and each lies in $U_{\infty}$ by step 1.1.

4.1step 3.1L2L7∎

By [L2] the index is constant on the connected component $U_{\infty}$ , and step 3.1 exhibits a point of $U_{\infty}$ where its value is $0$ ; hence $n (γ, p) = 0$ for every $p \in U_{\infty}$ , which by [L7] contains every connected unbounded subset of $C ∖ γ^{*}$ that meets it.

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Reversal negates and concatenation adds winding numbers

Statement

Reversal. Let $γ : [a, b] \to C$ be a closed complex contour and let $p \notin γ^{*}$ . Then the reversal $γ^{-} (t) = γ (a + b - t)$ is a closed complex contour with the same trace, and

$n (γ^{-}, p) = - n (γ, p) .$

Concatenation. Let $α, β : [0, 1] \to C$ be complex contours with $α (1) = β (0)$ . Then $α * β$ is a complex contour with trace $α^{*} \cup β^{*}$ , and for every $p \notin α^{*} \cup β^{*}$

$\int_{α * β} \frac{d z}{z - p} = \int_{α} \frac{d z}{z - p} + \int_{β} \frac{d z}{z - p} .$

If moreover $α$ and $β$ are themselves closed, then $α * β$ is closed and

$n (α * β, p) = n (α, p) + n (β, p) .$

The hypothesis $α (1) = β (0)$ is what makes the concatenation a contour, and closedness of $α * β$ is what makes its index defined; the integral identity needs neither $α$ nor $β$ to be closed.

Facts & Assumptions

Given: Closed complex contours where an index is asserted, composable complex contours where a concatenation is asserted, and a point off the traces involved.

[L1]

For a closed complex contour $γ$ and $p \notin γ^{*}$ , $n (γ, p) = (2 π i)^{- 1} \int_{γ} d z / (z - p)$ (The winding number of a closed contour about a point off its trace).

[L2]

For a rectifiable contour $γ$ , $\int_{γ^{-}} f d z = - \int_{γ} f d z$ ; for composable rectifiable contours $α, β$ , $\int_{α * β} f d z = \int_{α} f d z + \int_{β} f d z$ (Complex line integrals change sign under reversal and add under concatenation).

[L3]

A complex contour is a rectifiable path $γ : [a, b] \to C$ ; it is closed when $γ (a) = γ (b)$ ; its reversal is $γ^{-} (t) = γ (a + b - t)$ ; and for $α, β : [0, 1] \to C$ with $α (1) = β (0)$ the concatenation is $(α * β) (s) = α (2 s)$ for $s \leq \frac{1}{2}$ and $β (2 s - 1)$ for $s \geq \frac{1}{2}$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L4]

Arc length is unchanged by a monotone reparametrization (Arc length is invariant under every continuous surjective monotone reparametrization, including pauses and reversal).

[L5]

Arc length is additive across a split of the parameter interval, and a path is rectifiable exactly when both restrictions are (Arc length is additive across every subdivision point and decreases under restriction).

[L6]

For a rectifiable contour and a continuous integrand on its trace, the complex line integral exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L7]

The winding number of a closed complex contour about a point off its trace is an integer (The winding number of a closed contour is an integer).

Proof

technique · direct

1.1givenL3L4

The map $t \mapsto a + b - t$ is a decreasing continuous bijection of $[a, b]$ onto itself, so by [L3] and [L4] the reversal $γ^{-}$ is a path of the same length as $γ$ , hence rectifiable, and its trace is $γ ([a, b]) = γ^{*}$ ; it is closed because $γ^{-} (a) = γ (b) = γ (a) = γ^{-} (b)$ by [L3].

1.2givenL3L4L5

By [L3] the concatenation $α * β$ is continuous on $[0, 1]$ , its restrictions to $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ are monotone reparametrizations of $α$ and $β$ , so both are rectifiable by [L4] and $α * β$ is rectifiable by [L5]; its trace is $α^{*} \cup β^{*}$ by the two-piece formula.

1.3givenL3

If $α$ and $β$ are closed then $(α * β) (0) = α (0) = α (1) = β (0) = β (1) = (α * β) (1)$ by [L3], so $α * β$ is closed.

2.1step 1.1L1L2L6

With $p \notin γ^{*}$ the function $z \mapsto 1 / (z - p)$ is continuous on $γ^{*} = (γ^{-})^{*}$ , so all the integrals below exist by [L6]; applying the reversal identity of [L2] to it and dividing by $2 π i$ gives $n (γ^{-}, p) = - n (γ, p)$ through [L1].

2.2step 1.2L2L6

With $p \notin α^{*} \cup β^{*}$ the function $z \mapsto 1 / (z - p)$ is continuous on that union, so the concatenation identity of [L2] applies to it and gives the displayed additive formula, all three integrals existing by [L6].

3.1step 1.3step 2.1step 2.2L1L7∎

If $α$ and $β$ are closed, step 1.3 makes $α * β$ closed, so [L1] turns step 2.2 into $n (α * β, p) = n (α, p) + n (β, p)$ ; all three values are integers by [L7], consistently with the identity.

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A circle traversed $k$ times has winding number $k$ inside and $0$ outside

Statement

Let $a \in C$ , let $r > 0$ , let $k \in Z$ and put

$γ_{k} (t) = a + r \exp (i k t), t \in [0, 2 π] .$

Then $γ_{k}$ is a closed complex contour and

$n (γ_{k}, z) = k for ∣ z - a ∣ < r, n (γ_{k}, z) = 0 for ∣ z - a ∣ > r .$

For $k \neq 0$ the trace of $γ_{k}$ is the circle ${z : ∣ z - a ∣ = r}$ . For $k = 0$ the contour is the constant path at $a + r$ and its trace is ${a + r}$ ; the two displayed formulas still hold, both values being $0$ , and both regions still lie off the trace.

Facts & Assumptions

Given: A point $a \in C$ , a real $r > 0$ and an integer $k$ .

[L1]

For a closed complex contour $γ$ and $p \notin γ^{*}$ , $n (γ, p) = (2 π i)^{- 1} \int_{γ} d z / (z - p)$ (The winding number of a closed contour about a point off its trace).

[L2]

The winding number of a closed complex contour about a point off its trace is an integer (The winding number of a closed contour is an integer).

[L3]

The index is constant on every connected component of the complement of the trace (The winding number is constant on each connected component of the complement of the trace).

[L4]

The complement of the trace of a closed complex contour has exactly one unbounded connected component, and the index vanishes there (The winding number vanishes on the unbounded component of the complement of the trace); for a compact $K$ the complement $C ∖ K$ has exactly one unbounded component and every other component is bounded (The complement of a compact plane set has exactly one unbounded connected component).

[L5]

For $c \in C$ and $R > 0$ the set ${z : ∣ z - c ∣ > R}$ is path-connected and connected (The exterior of a closed disc in the plane is path-connected).

[L6]

For a complex contour $γ : [a, b] \to C$ , a point $p \notin γ^{*}$ and a continuous logarithm $λ$ of $γ - p$ along $γ$ , $\int_{γ} d z / (z - p) = λ (b) - λ (a)$ (The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm); such a $λ$ is a continuous map with $\exp (λ (t)) = γ (t) - p$ throughout (Continuous logarithms and continuous arguments along a contour).

[L7]

For a positively oriented circle $σ (t) = a + r \exp (i t)$ with $r > 0$ , $(2 π i)^{- 1} \int_{σ} d z / (z - a) = 1$ (The normalized integral around a positively oriented circle centred at a is 1).

[L8]

$\exp (z + w) = \exp z \exp w$ for all complex $z, w$ , and $\exp (x + 0 i) = e^{x}$ for real $x$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential); for real $x, y$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ and $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L9]

$\ker (\exp) = 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L10]

$t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the unit circle $S^{1}$ ( $t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle), and $\cos$ and $\sin$ have fundamental period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[L11]

$\cos$ and $\sin$ are differentiable on $R$ with $\cos^{'} = - \sin$ and $\sin^{'} = \cos$ (The derivatives of sine and cosine are cosine and minus sine).

[L12]

A continuous path that is differentiable with a continuous derivative on each piece of a partition is rectifiable (A continuous piecewise- $C^{1}$ path is rectifiable and its length is the sum of the speed integrals over its pieces).

[L13]

For $x > 0$ , $\log x$ is the unique real $y$ with $\exp y = x$ (The natural logarithm as the inverse of the exponential function).

[L14]

$B (x, ρ) = {y : d (x, y) < ρ}$ (Open ball, closed ball and sphere in a metric space); a set is convex when it contains the segment between any two of its points (A convex subset of $R^{m}$ contains every line segment between two of its points); a subset joined by paths inside it is path-connected (Paths, path-connected spaces and path components) and hence connected (Every path-connected space is connected, and every path component lies inside a component).

[L15]

Distinct components are disjoint and every connected subset containing a point lies inside that point's component (The components of a space are its maximal connected subsets, they partition it, and each of them is closed, Connected components, quasicomponents, and totally disconnected spaces).

[L16]

A subset of a metric space is bounded when it is empty or contained in some ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L17]

[L18]

The integers form a commutative ring (The integers form a commutative ring).

Proof

technique · direct

1.1givenL8L9L11L12

By [L8] one has $γ_{k} (t) = a + r \cos (k t) + i r \sin (k t)$ , which by [L11] is differentiable in $t$ with the continuous derivative $- k r \sin (k t) + i k r \cos (k t)$ , so $γ_{k}$ is rectifiable by [L12]; and $γ_{k} (2 π) = a + r \exp (2 π i k) = a + r = γ_{k} (0)$ by [L9], so $γ_{k}$ is a closed complex contour.

1.2givenL6L8L13L17

Since $∣ γ_{k} (t) - a ∣ = r ∣ \exp (i k t) ∣ = r > 0$ by [L8] and [L17], the centre $a$ lies off the trace, and $λ (t) = \log r + i k t$ is a continuous map on $[0, 2 π]$ with $\exp (λ (t)) = e^{\log r} \exp (i k t) = r \exp (i k t) = γ_{k} (t) - a$ by [L8] and [L13]; so $λ$ is a continuous logarithm of $γ_{k} - a$ along $γ_{k}$ in the sense of [L6].

1.3givenL8L10L17

The trace of $γ_{k}$ is ${a + r \exp (i k t) : 0 \leq t \leq 2 π}$ . If $k = 0$ this is ${a + r}$ . If $k \neq 0$ then ${k t : 0 \leq t \leq 2 π}$ is a closed interval of length $2 π ∣ k ∣ \geq 2 π$ , so by the periodicity and surjectivity in [L10] the values $\exp (i k s)$ run over the whole unit circle, and the trace is ${z : ∣ z - a ∣ = r}$ by [L8] and [L17]. In both cases the trace is contained in ${z : ∣ z - a ∣ = r}$ .

2.1step 1.1step 1.2L1L2L6L7L18

By [L6] and step 1.2, $\int_{γ_{k}} d z / (z - a) = λ (2 π) - λ (0) = 2 π i k$ , so $n (γ_{k}, a) = k$ by [L1]; for $k = 1$ this is the published normalisation [L7], and by [L2] and [L18] the value is an integer, as it must be.

2.2step 1.3L14L17

The disc $D = B (a, r)$ is convex by [L14] and [L17], hence path-connected along segments and therefore connected; step 1.3 puts the trace in ${z : ∣ z - a ∣ = r}$ , which is disjoint from $D$ , so $D \subseteq C ∖ γ_{k}^{*}$ and $a \in D$ .

2.3step 1.3L4L5L15L16L17

The set $E = {z : ∣ z - a ∣ > r}$ is connected by [L5] and is disjoint from the trace by step 1.3; it is unbounded by [L16] and [L17], so by [L15] it lies in a single component of $C ∖ γ_{k}^{*}$ , and that component is unbounded, hence is the unique unbounded one of [L4].

3.1step 2.1step 2.2L3L15

By [L3] the index is constant on the component of $C ∖ γ_{k}^{*}$ containing $D$ , and $D$ is a connected subset of that complement containing $a$ , so by [L15] it lies in one component; hence $n (γ_{k}, z) = n (γ_{k}, a) = k$ for every $z \in D$ , that is for $∣ z - a ∣ < r$ .

4.1step 3.1step 2.3L4∎

By [L4] the index vanishes on that unique unbounded component, so $n (γ_{k}, z) = 0$ for every $z \in E$ by step 2.3, while step 3.1 gives the value $k$ on $∣ z - a ∣ < r$ ; when $k = 0$ both regions still lie off the single-point trace of step 1.3 and both values are $0$ .

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Complex chains, their traces, and cycles

Definition

Let $Ω \subseteq C$ be open. A complex chain in $Ω$ is a finite list

$Γ = ((m_{0}, γ_{0}), \dots, (m_{r - 1}, γ_{r - 1})), r \in N,$

each $m_{k}$ an integer (The integers as equivalence classes of pairs of naturals) and each $γ_{k} : [a_{k}, b_{k}] \to C$ a complex contour (Rectifiable complex contours, reversal, concatenation, closedness, and orientation) whose trace lies in $Ω$ . It is written $\sum_{k < r} m_{k} γ_{k}$ . The list of length $r = 0$ is the empty chain.

Its trace is the set

$Γ^{*} : = ⋃ {γ_{k}^{*} : k < r and m_{k} \neq 0},$

a subset of $Ω$ ; a term with $m_{k} = 0$ contributes nothing to it.

Its boundary is the function $\partial Γ : C \to Z$ given by

$\partial Γ (q) : = \sum_{\begin{array}{c} k < r \\ γ_{k} (b_{k}) = q \end{array}} m_{k} - \sum_{\begin{array}{c} k < r \\ γ_{k} (a_{k}) = q \end{array}} m_{k},$

each sum being a finite sum over a subset of ${k : k < r}$ (A finite sum in a commutative monoid indexed by an arbitrary finite set, The cardinality $∣ A ∣$ of a finite set), and the subtraction is that of the commutative ring $Z$ (The integers form a commutative ring). The chain $Γ$ is a cycle when $\partial Γ (q) = 0$ for every $q \in C$ .

Sum and negation. For chains $Γ_{1}$ and $Γ_{2}$ in $Ω$ , the sum $Γ_{1} + Γ_{2}$ is the concatenated list, and $- Γ$ is the list with every coefficient replaced by $- m_{k}$ and every contour unchanged. Write $Γ_{1} - Γ_{2} : = Γ_{1} + (- Γ_{2})$ . The reversal $Γ^{-}$ is the list with every coefficient unchanged and every contour replaced by its reversal; the additive inverse $- m_{k}$ is taken in $Z$ (The integers form a commutative ring).

Remarks

A chain is a list, and equality of chains is equality of lists. No free abelian group on the set of contours is introduced here, and no result on this page asserts that two differently presented chains are equal: every statement below is about a given list, and the operations above produce lists. This is a deliberate departure from the presentations that define a chain as a group element, and it is what removes the obligation to say when two chains coincide.

Which lists are cycles. The empty chain is a cycle, both boundary sums being empty and hence $0$ (A finite sum in a commutative monoid indexed by an arbitrary finite set). A list all of whose contours are closed is a cycle: for such a $γ_{k}$ the two endpoint values coincide, so $m_{k}$ enters the sum at $q = γ_{k} (a_{k})$ once positively and once negatively and cancels there, and enters neither sum at any other point. In particular a single closed contour, taken as the list of length $1$ with $m_{0} = 1$ , is a cycle. Terms with $m_{k} = 0$ add $0$ to both sums and so never affect $\partial Γ$ .

Cycles are more general than lists of closed contours. The condition is that the endpoints cancel after the coefficients are counted, not that each piece closes up: two contours with the same initial point and the same terminal point, carried with coefficients $+ 1$ and $- 1$ , form a cycle although neither is closed. The distinction is what the integral results below actually use: the vanishing of the boundary function is exactly the hypothesis under which the integral of a continuous derivative over $Γ$ is zero.

Ambient set. A chain is a chain in $Ω$ ; the same list is a chain in every open set containing all the $γ_{k}^{*}$ , and in particular in $C$ . When a nonempty connected $Ω$ is wanted it is called a complex domain (A complex domain is a nonempty connected open subset of $C$ ). Finite sums of integers and of complex numbers are finite sums in their additive commutative monoids, as in A finite sum in a commutative monoid indexed by an arbitrary finite set.

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Integration over a complex chain and the index of a chain

Definition

Let $Γ = \sum_{k < r} m_{k} γ_{k}$ be a complex chain (Complex chains, their traces, and cycles) with trace $Γ^{*}$ , and let $f$ be continuous on $Γ^{*}$ . The integral of $f$ over $Γ$ is

$\int_{Γ} f (z) d z : = \sum_{\begin{array}{c} k < r \\ m_{k} \neq 0 \end{array}} m_{k} \int_{γ_{k}} f (z) d z,$

a finite sum (A finite sum in a commutative monoid indexed by an arbitrary finite set) of the complex line integrals of The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral. Each summand exists: for $k$ with $m_{k} \neq 0$ the trace $γ_{k}^{*}$ is contained in $Γ^{*}$ , so $f$ is continuous on it, and $γ_{k}$ is rectifiable, so Continuous integrands have complex and absolute line integrals along every rectifiable path applies. Terms with $m_{k} = 0$ are omitted, so no integral of $f$ over a contour outside the trace is required. The empty chain, and any chain all of whose coefficients vanish, give $\int_{Γ} f d z = 0$ .

For $p \in C ∖ Γ^{*}$ , the index of $Γ$ about $p$ is

$n (Γ, p) : = \frac{1}{2 π i} \int_{Γ} \frac{d z}{z - p} .$

This is defined: $z \mapsto 1 / (z - p)$ is complex differentiable, hence continuous, on $C ∖ {p} \supseteq Γ^{*}$ by Linearity, product, reciprocal, and quotient rules for complex derivatives and Complex differentiability at a point implies continuity there.

Remarks

The notation is consistent with the single-contour case. If $r = 1$ , $m_{0} = 1$ and $γ_{0}$ is closed, then $Γ^{*} = γ_{0}^{*}$ , the sum has the one term $\int_{γ_{0}} f d z$ , and $n (Γ, p)$ is the winding number $n (γ_{0}, p)$ of The winding number of a closed contour about a point off its trace for every $p \notin γ_{0}^{*}$ . So writing $n$ for both costs no ambiguity.

Linearity in the integrand is inherited termwise from Complex line integrals are linear in the integrand, finite sums in the additive commutative monoid of $C$ (A finite sum in a commutative monoid indexed by an arbitrary finite set), and distributivity in the complex field ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ): for $f, g$ continuous on $Γ^{*}$ and $α, β \in C$ , $\int_{Γ} (α f + β g) d z = α \int_{Γ} f d z + β \int_{Γ} g d z$ .

The index is not defined on the trace. For $p \in Γ^{*}$ the integrand is undefined at $z = p$ , and no value is assigned; every statement about $n (Γ, \cdot)$ below carries the hypothesis $p \notin Γ^{*}$ .

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Chain integration and the index are additive in the chain, and reverse with it

Statement

Let $Γ$ , $Γ_{1}$ , $Γ_{2}$ be complex chains (Complex chains, their traces, and cycles). Then:

$(Γ_{1} + Γ_{2})^{*} = Γ_{1}^{*} \cup Γ_{2}^{*}$ , $(- Γ)^{*} = Γ^{*}$ and $(Γ^{-})^{*} = Γ^{*}$ ;
$\partial (Γ_{1} + Γ_{2}) = \partial Γ_{1} + \partial Γ_{2}$ , $\partial (- Γ) = - \partial Γ$ and $\partial (Γ^{-}) = - \partial Γ$ as functions on $C$ ; consequently a sum of cycles is a cycle, and the negative and the reversal of a cycle are cycles;
for $f$ continuous on $Γ_{1}^{*} \cup Γ_{2}^{*}$ , $\int_{Γ_{1} + Γ_{2}} f d z = \int_{Γ_{1}} f d z + \int_{Γ_{2}} f d z,$ and for $f$ continuous on $Γ^{*}$ , $\int_{- Γ} f d z = - \int_{Γ} f d z, \int_{Γ^{-}} f d z = - \int_{Γ} f d z;$
for $p \notin Γ_{1}^{*} \cup Γ_{2}^{*}$ , $n (Γ_{1} + Γ_{2}, p) = n (Γ_{1}, p) + n (Γ_{2}, p)$ , and for $p \notin Γ^{*}$ , $n (- Γ, p) = - n (Γ, p) = n (Γ^{-}, p)$ ;
the empty chain is a cycle, its integral of every function is $0$ , and its index is $0$ at every point of $C$ .

Facts & Assumptions

Given: Complex chains $Γ = \sum_{k < r} m_{k} γ_{k}$ , $Γ_{1} = \sum_{k < r_{1}} m_{k}^{1} γ_{k}^{1}$ and $Γ_{2} = \sum_{k < r_{2}} m_{k}^{2} γ_{k}^{2}$ , and integrands continuous on the traces named in each clause.

[L1]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ and, for $p \notin Γ^{*}$ , $n (Γ, p) = (2 π i)^{- 1} \int_{Γ} d z / (z - p)$ (Integration over a complex chain and the index of a chain).

[L2]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ ; its trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ ; its boundary is $\partial Γ (q) = \sum {m_{k} : γ_{k} (b_{k}) = q} - \sum {m_{k} : γ_{k} (a_{k}) = q}$ ; it is a cycle when that vanishes identically; $Γ_{1} + Γ_{2}$ is list concatenation, $- Γ$ negates every coefficient, and $Γ^{-}$ reverses every contour (Complex chains, their traces, and cycles).

[L3]

For a rectifiable contour $γ$ , $\int_{γ^{-}} f d z = - \int_{γ} f d z$ (Complex line integrals change sign under reversal and add under concatenation); the reversal of a closed contour is a closed contour with the same trace and $n (γ^{-}, p) = - n (γ, p)$ (Reversal negates and concatenation adds winding numbers).

[L4]

For continuous $f, g$ on the trace of a rectifiable contour and $α, β \in C$ , $\int_{γ} (α f + β g) d z = α \int_{γ} f d z + β \int_{γ} g d z$ (Complex line integrals are linear in the integrand).

[L5]

Finite sums in the additive commutative monoid of $C$ are additive, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L6]

For disjoint finite index sets $S, T$ and $a : S \cup T \to M$ in a commutative monoid, $\sum_{u \in S \cup T} a_{u} = \sum_{s \in S} a_{s} + \sum_{t \in T} a_{t}$ (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule); a sum over a finite index set is well posed and the sum over the empty set is $0$ (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[L7]

For a rectifiable contour and a continuous integrand on its trace, the complex line integral exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L8]

The reversal of $γ : [a, b] \to C$ is $γ^{-} (t) = γ (a + b - t)$ , so its endpoints are exchanged (Rectifiable complex contours, reversal, concatenation, closedness, and orientation), and arc length is unchanged by a monotone reparametrization (Arc length is invariant under every continuous surjective monotone reparametrization, including pauses and reversal).

Proof

technique · direct

1.1givenL2L8

The list $Γ_{1} + Γ_{2}$ has as its terms exactly the terms of $Γ_{1}$ followed by those of $Γ_{2}$ , so the set of indices with nonzero coefficient splits as a disjoint union and [L2] gives $(Γ_{1} + Γ_{2})^{*} = Γ_{1}^{*} \cup Γ_{2}^{*}$ . Negating a coefficient does not change whether it is zero, and reversing a contour does not change its trace by [L8], so $(- Γ)^{*} = (Γ^{-})^{*} = Γ^{*}$ . This is claim 1.

1.2givenL2L5L6L8

For each $q$ , the two defining sums of $\partial (Γ_{1} + Γ_{2}) (q)$ run over the disjoint union of the corresponding index sets for $Γ_{1}$ and $Γ_{2}$ , so [L6] splits each of them and gives $\partial (Γ_{1} + Γ_{2}) (q) = \partial Γ_{1} (q) + \partial Γ_{2} (q)$ . Replacing every $m_{k}$ by $- m_{k}$ multiplies both sums by $- 1$ by [L5], giving $\partial (- Γ) = - \partial Γ$ ; and by [L8] the reversal exchanges the roles of the two endpoint sums, giving $\partial (Γ^{-}) = - \partial Γ$ . Hence if $\partial Γ_{1}$ and $\partial Γ_{2}$ vanish identically so does $\partial (Γ_{1} + Γ_{2})$ , and likewise for $- Γ$ and $Γ^{-}$ ; this is claim 2.

1.3L1L2L6

The empty chain has empty trace, both its boundary sums are empty and therefore $0$ by [L6], and its integral is the empty sum, which is $0$ by [L1] and [L6]; hence its index is $0$ at every $p \in C$ , all of which lie off its empty trace. This is claim 5.

2.1step 1.1L1L6L7

Let $f$ be continuous on $Γ_{1}^{*} \cup Γ_{2}^{*}$ , which by step 1.1 is the trace of $Γ_{1} + Γ_{2}$ , so all three integrals exist by [L1] and [L7]. The defining sum for $\int_{Γ_{1} + Γ_{2}} f d z$ runs over the disjoint union of the two index sets, so [L6] splits it into $\int_{Γ_{1}} f d z + \int_{Γ_{2}} f d z$ .

3.1step 1.1step 2.1L1L3L5

Let $f$ be continuous on $Γ^{*}$ . Replacing each $m_{k}$ by $- m_{k}$ multiplies each summand of [L1] by $- 1$ , so [L5] gives $\int_{- Γ} f d z = - \int_{Γ} f d z$ ; and replacing each $γ_{k}$ by $γ_{k}^{-}$ negates each $\int_{γ_{k}} f d z$ by [L3], so $\int_{Γ^{-}} f d z = - \int_{Γ} f d z$ as well. Together with step 2.1 this is claim 3.

4.1step 1.3step 2.1step 3.1L1L3L4∎

Applying steps 2.1 and 3.1 to $f (z) = 1 / (z - p)$ , which is continuous on the traces involved whenever $p$ lies off them, and dividing by $2 π i$ using [L4] gives claim 4: $n (Γ_{1} + Γ_{2}, p) = n (Γ_{1}, p) + n (Γ_{2}, p)$ and $n (- Γ, p) = - n (Γ, p) = n (Γ^{-}, p)$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The index of a cycle about a point off its trace is an integer

Statement

Let $Γ = \sum_{k < r} m_{k} γ_{k}$ be a complex chain which is a cycle, that is $\partial Γ$ vanishes identically (Complex chains, their traces, and cycles), and let $p \in C$ with $p \notin Γ^{*}$ . Then

$n (Γ, p) \in Z .$

The individual contours $γ_{k}$ need not be closed; what is used is that the endpoint counts cancel. The empty cycle gives $n (Γ, p) = 0$ .

Facts & Assumptions

Given: A complex chain $Γ = \sum_{k < r} m_{k} γ_{k}$ with $γ_{k} : [a_{k}, b_{k}] \to C$ , whose boundary function vanishes identically, and a point $p \notin Γ^{*}$ .

[L1]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ , its trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ , its boundary is $\partial Γ (q) = \sum {m_{k} : γ_{k} (b_{k}) = q} - \sum {m_{k} : γ_{k} (a_{k}) = q}$ , and it is a cycle when that function vanishes identically (Complex chains, their traces, and cycles).

[L2]

[L3]

[L4]

For a complex contour $γ$ and $p \notin γ^{*}$ there is a continuous logarithm of $γ - p$ along $γ$ (Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$ ), a continuous $λ$ with $\exp (λ (t)) = γ (t) - p$ throughout (Continuous logarithms and continuous arguments along a contour).

[L5]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L6]

The complex exponential maps $C$ onto $C ∖ {0}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

[L7]

Finite sums in the additive commutative group of $C$ are additive and telescope, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L8]

For disjoint finite index sets $S, T$ , $\sum_{u \in S \cup T} a_{u} = \sum_{s \in S} a_{s} + \sum_{t \in T} a_{t}$ (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule); sums over finite index sets are well posed and the empty sum is $0$ (A finite sum in a commutative monoid indexed by an arbitrary finite set, The cardinality $∣ A ∣$ of a finite set).

[L9]

A natural-number-indexed finite list of nonempty sets admits a choice function, provably in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[L10]

The integers form a commutative ring (The integers form a commutative ring).

Proof

technique · direct

1.1givenL1L8

Write $K = {k < r : m_{k} \neq 0}$ and let $Q = {γ_{k} (a_{k}) : k \in K} \cup {γ_{k} (b_{k}) : k \in K}$ , a finite subset of $Γ^{*}$ by [L1]; in particular $q - p \neq 0$ for every $q \in Q$ , since $p \notin Γ^{*}$ .

1.2givenL4L6L9

For each $k \in K$ the point $p$ lies off $γ_{k}^{*} \subseteq Γ^{*}$ , so [L4] provides a continuous logarithm $λ_{k}$ of $γ_{k} - p$ along $γ_{k}$ ; finitely many such choices are made, which is legitimate in ZF by [L9]. Likewise [L6] and [L9] provide, for each $q \in Q$ , a complex number $μ (q)$ with $\exp (μ (q)) = q - p$ .

2.1step 1.2L2L3

By [L2] and [L3], $n (Γ, p) = \frac{1}{2 π i} \sum_{k \in K} m_{k} (λ_{k} (b_{k}) - λ_{k} (a_{k}))$ .

2.2step 1.1step 1.2L5

Fix $k \in K$ . Since $\exp (λ_{k} (a_{k})) = γ_{k} (a_{k}) - p = \exp (μ (γ_{k} (a_{k})))$ and similarly at $b_{k}$ , [L5] gives integers $u_{k}, v_{k}$ with $λ_{k} (a_{k}) = μ (γ_{k} (a_{k})) + 2 π i u_{k}$ and $λ_{k} (b_{k}) = μ (γ_{k} (b_{k})) + 2 π i v_{k}$ ; hence $λ_{k} (b_{k}) - λ_{k} (a_{k}) = μ (γ_{k} (b_{k})) - μ (γ_{k} (a_{k})) + 2 π i (v_{k} - u_{k})$ .

3.1step 2.1step 2.2L7L10

Substituting step 2.2 into step 2.1 and using [L7], one gets $n (Γ, p) = \frac{S}{2 π i} + \sum_{k \in K} m_{k} (v_{k} - u_{k})$ , where $S = \sum_{k \in K} m_{k} μ (γ_{k} (b_{k})) - \sum_{k \in K} m_{k} μ (γ_{k} (a_{k}))$ . The second summand is an integer by [L10].

4.1step 1.1step 3.1L1L7L8

The index set $K$ is the disjoint union over $q \in Q$ of ${k \in K : γ_{k} (b_{k}) = q}$ , so [L8] and [L7] give $\sum_{k \in K} m_{k} μ (γ_{k} (b_{k})) = \sum_{q \in Q} μ (q) \sum {m_{k} : k \in K, γ_{k} (b_{k}) = q}$ , and likewise with $a_{k}$ in place of $b_{k}$ ; since a term with $m_{k} = 0$ contributes $0$ to the boundary sums of [L1], subtracting the two gives $S = \sum_{q \in Q} μ (q) \partial Γ (q)$ , which is $0$ because $Γ$ is a cycle.

5.1step 3.1step 4.1L8L10∎

Step 4.1 makes $S = 0$ , so step 3.1 gives $n (Γ, p) = \sum_{k \in K} m_{k} (v_{k} - u_{k})$ , an integer by [L10]. For the empty cycle $K$ is empty and the sum of step 2.1 is $0$ by [L8], giving $n (Γ, p) = 0$ .

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The integral of a continuous derivative over a cycle is zero

Statement

Let $Γ = \sum_{k < r} m_{k} γ_{k}$ be a complex chain which is a cycle, let $V \subseteq C$ be open with $Γ^{*} \subseteq V$ , and let $F$ be a primitive on $V$ of a continuous $f$ , so that $F^{'} = f$ is continuous (A primitive of a complex function on an open set). Then

$\int_{Γ} f (z) d z = 0.$

The hypothesis used is that the boundary function of $Γ$ vanishes, which is weaker than requiring every $γ_{k}$ to be closed.

Facts & Assumptions

Given: A cycle $Γ = \sum_{k < r} m_{k} γ_{k}$ with $γ_{k} : [a_{k}, b_{k}] \to C$ , an open $V \supseteq Γ^{*}$ , and a primitive $F$ on $V$ of a continuous $f$ with $F^{'} = f$ continuous.

[L1]

[L2]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ (Integration over a complex chain and the index of a chain).

[L3]

[L4]

[L5]

[L6]

A primitive of $f$ on $V$ is a holomorphic $F$ with $F^{'} = f$ on $V$ (A primitive of a complex function on an open set).

Proof

technique · direct

1.1givenL1L5L6

Write $K = {k < r : m_{k} \neq 0}$ and $Q = {γ_{k} (a_{k}) : k \in K} \cup {γ_{k} (b_{k}) : k \in K}$ , a finite subset of $Γ^{*} \subseteq V$ by [L1]; so $F$ is defined at every point of $Q$ .

1.2givenL1L3L6

For every $k \in K$ the trace $γ_{k}^{*}$ lies in the open set $V$ on which $F$ is a primitive of the continuous $f$ with continuous $F^{'}$ , so [L3] gives $\int_{γ_{k}} f d z = F (γ_{k} (b_{k})) - F (γ_{k} (a_{k}))$ .

2.1step 1.2L2L4

By [L2] and step 1.2, $\int_{Γ} f d z = \sum_{k \in K} m_{k} F (γ_{k} (b_{k})) - \sum_{k \in K} m_{k} F (γ_{k} (a_{k}))$ , using [L4] to split the sum.

3.1step 1.1step 2.1L1L4L5

The index set $K$ is the disjoint union over $q \in Q$ of ${k \in K : γ_{k} (b_{k}) = q}$ , so [L5] and [L4] give $\sum_{k \in K} m_{k} F (γ_{k} (b_{k})) = \sum_{q \in Q} F (q) \sum {m_{k} : k \in K, γ_{k} (b_{k}) = q}$ , and likewise with $a_{k}$ in place of $b_{k}$ ; a term with $m_{k} = 0$ contributes $0$ to the boundary sums of [L1], so subtracting gives $\int_{Γ} f d z = \sum_{q \in Q} F (q) \partial Γ (q)$ .

4.1step 3.1L1L5∎

Every $\partial Γ (q)$ vanishes because $Γ$ is a cycle, so the sum of step 3.1 is $0$ , whence $\int_{Γ} f d z = 0$ ; the same conclusion holds for the empty cycle, whose defining sum is empty and therefore $0$ by [L5].

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The index of a cycle is locally constant off its trace and vanishes far from it

Statement

Let $Γ = \sum_{k < r} m_{k} γ_{k}$ be a complex chain which is a cycle, and put $M (Γ) = \sum_{k < r, m_{k} \neq 0} ∣ m_{k} ∣ L (γ_{k})$ . Then the trace $Γ^{*}$ is compact, $C ∖ Γ^{*}$ is open, and:

for $p_{0} \notin Γ^{*}$ with $d = \inf {∣ w - p_{0} ∣ : w \in Γ^{*}}$ and $∣ p - p_{0} ∣ < d / 2$ , $∣ n (Γ, p) - n (Γ, p_{0}) ∣ \leq \frac{M (Γ) ∣ p - p_{0} ∣}{π d^{2}},$ so $n (Γ, \cdot)$ is continuous on $C ∖ Γ^{*}$ ;
$n (Γ, \cdot)$ is constant on every connected component of $C ∖ Γ^{*}$ , and each such component is open, so the index is locally constant;
there is $R > 0$ with $n (Γ, p) = 0$ for every $p$ with $∣ p ∣ > R$ .

Consequently $Ω_{0} : = {p \in C ∖ Γ^{*} : n (Γ, p) = 0}$ is open and contains ${p : ∣ p ∣ > R}$ .

If $Γ^{*} = \emptyset$ then $d$ is the infimum of the empty set and clause 1 is not asserted; in that case $n (Γ, p) = 0$ for every $p \in C$ and clauses 2 and 3 hold with $Ω_{0} = C$ .

Facts & Assumptions

Given: A cycle $Γ = \sum_{k < r} m_{k} γ_{k}$ ; the plane carries the Euclidean metric of $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

The index of a cycle about a point off its trace is an integer (The index of a cycle about a point off its trace is an integer).

[L2]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ and its trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ (Complex chains, their traces, and cycles).

[L3]

[L4]

[L5]

[L6]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded); the continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); a finite union of compact subsets is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact); a closed bounded interval is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L7]

The connected component $C (x)$ is the union of all connected subsets containing $x$ (Connected components, quasicomponents, and totally disconnected spaces), and every component of an open subset of $R^{n}$ is open and polygonally connected (Every connected component of an open subset of $R^{n}$ is open and polygonally connected).

[L8]

[L9]

A set is closed exactly when its complement is open, and a set is open exactly when each of its points admits a ball inside it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space); a subset is bounded when it is empty or lies inside some ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L10]

A nonempty subset of $R$ bounded below has a greatest lower bound (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

[L11]

[L12]

[L13]

The integers form an ordered commutative ring and are discrete in $R$ ; in particular the only integer of modulus below $1$ is $0$ , and if $m < n$ then $m + \frac{1}{2}$ lies strictly between them and is not an integer (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

Proof

technique · direct

1.1givenL2L6L9

Each $γ_{k}^{*}$ is the continuous image of a compact interval, hence compact by [L6], so the trace $Γ^{*}$ of [L2] is a finite union of compact sets and is compact by [L6], closed and bounded by [L6], and its complement is open by [L9].

2.1step 1.1L9L10L11algebra

Suppose $Γ^{*} \neq \emptyset$ , fix $p_{0} \notin Γ^{*}$ and put $d = \inf {∣ w - p_{0} ∣ : w \in Γ^{*}}$ , which exists by [L10] and is positive because the complement of the closed set $Γ^{*}$ is open, so some ball $B (p_{0}, ε)$ misses $Γ^{*}$ and $d \geq ε$ by [L9]. For $∣ p - p_{0} ∣ < d / 2$ and $z \in Γ^{*}$ one has $∣ z - p_{0} ∣ \geq d$ and $∣ z - p ∣ > d / 2 > 0$ by [L11], so $p \notin Γ^{*}$ and $∣ \frac{1}{z - p} - \frac{1}{z - p_{0}} ∣ = ∣ \frac{p - p_{0}}{(z - p) (z - p_{0})} ∣ \leq 2 ∣ p - p_{0} ∣ / d^{2}$ .

2.2step 1.1L3L4L9L11L12

Let $R_{0} > 0$ satisfy $Γ^{*} \subseteq {z : ∣ z ∣ \leq R_{0}}$ , available from the boundedness in step 1.1 and [L9]. For $∣ p ∣ > R_{0}$ and $z \in Γ^{*}$ , [L11] gives $∣ z - p ∣ \geq ∣ p ∣ - R_{0} > 0$ , so [L3], [L4] and [L12] give $∣ n (Γ, p) ∣ \leq M (Γ) / (2 π (∣ p ∣ - R_{0}))$ .

3.1step 2.1L3L4L5L11L12

For $k$ with $m_{k} \neq 0$ the trace $γ_{k}^{*}$ lies in $Γ^{*}$ , so the bound of step 2.1 holds on it and [L4] gives $∣ \int_{γ_{k}} (\frac{1}{z - p} - \frac{1}{z - p_{0}}) d z ∣ \leq 2 ∣ p - p_{0} ∣ L (γ_{k}) / d^{2}$ ; combining the terms with [L3], [L5], [L11] and [L12] yields $∣ n (Γ, p) - n (Γ, p_{0}) ∣ \leq M (Γ) ∣ p - p_{0} ∣ / (π d^{2})$ , which is clause 1 and makes $n (Γ, \cdot)$ continuous at $p_{0}$ .

3.2step 2.2L1L13

Take $R = R_{0} + M (Γ) / (2 π) + 1$ . For $∣ p ∣ > R$ step 2.2 gives $∣ n (Γ, p) ∣ < 1$ , and $n (Γ, p)$ is an integer by [L1], so it is $0$ by [L13]; this is clause 3.

4.1step 1.1step 3.1L1L7L8L13

Let $C$ be a connected component of $C ∖ Γ^{*}$ . By [L7], applied to the open set of step 1.1, the component $C$ is open and polygonally connected, hence connected; by [L1] the index is integer-valued and by step 3.1 it is continuous, so [L8] makes its image on $C$ a connected subset of $R$ . That image lies in $Z$ , so [L13] forces it to be a single point. This is clause 2.

5.1step 3.2step 4.1L3L7∎

By step 4.1 the set $Ω_{0}$ is a union of components of the open set $C ∖ Γ^{*}$ , each open by [L7], hence open; and it contains ${p : ∣ p ∣ > R}$ by step 3.2. If $Γ^{*} = \emptyset$ then every $m_{k}$ is zero or $r = 0$ , so $\int_{Γ} f d z = 0$ for every $f$ by [L3] and $n (Γ, p) = 0$ for every $p \in C$ , giving $Ω_{0} = C$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Null-homologous cycles and homologous cycles in an open set

Definition

Let $Ω \subseteq C$ be open and let $Γ$ be a complex chain which is a cycle and whose trace lies in $Ω$ (Complex chains, their traces, and cycles).

$Γ$ is null-homologous in $Ω$ when

$n (Γ, p) = 0 for every p \in C ∖ Ω,$

the index being that of Integration over a complex chain and the index of a chain; the values are defined because $Γ^{*} \subseteq Ω$ , so every $p \notin Ω$ lies off the trace, and they are integers by The index of a cycle about a point off its trace is an integer.

Two cycles $Γ_{1}, Γ_{2}$ with traces in $Ω$ are homologous in $Ω$ when $Γ_{1} - Γ_{2}$ is null-homologous in $Ω$ . By Chain integration and the index are additive in the chain, and reverse with it the chain $Γ_{1} - Γ_{2}$ is again a cycle with trace inside $Γ_{1}^{*} \cup Γ_{2}^{*} \subseteq Ω$ , and its index at a point $p$ off that union is $n (Γ_{1}, p) - n (Γ_{2}, p)$ ; so the condition says exactly that

$n (Γ_{1}, p) = n (Γ_{2}, p) for every p \in C ∖ Ω .$

Remarks

Both notions depend on $Ω$ , not on the cycle alone. The same cycle can be null-homologous in one open set and not in another: enlarging $Ω$ removes points from $C ∖ Ω$ and so weakens the requirement. Every statement below that uses these words names the open set it uses them in, and $Ω$ is not omitted anywhere.

Null-homologous does not mean equal to the empty chain. It is a condition on the numbers $n (Γ, p)$ for $p$ outside $Ω$ , and a cycle with a large trace can satisfy it. In particular, being homologous is a relation between two cycles and never an assertion that the two lists coincide; chains here are lists and equality of chains is equality of lists.

Taking $Ω = C$ makes the condition vacuous, since $C ∖ C$ is empty, so every cycle is null-homologous in the plane. The content of the notion appears when $Ω$ omits points, and it is those omitted points that the index has to ignore. When a nonempty connected $Ω$ is wanted it is called a complex domain (A complex domain is a nonempty connected open subset of $C$ ).

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Homologically simply connected complex domains

Definition

A complex domain $Ω$ (A complex domain is a nonempty connected open subset of $C$ ) is homologically simply connected when every complex chain which is a cycle and whose trace lies in $Ω$ (Complex chains, their traces, and cycles) is null-homologous in $Ω$ (Null-homologous cycles and homologous cycles in an open set); equivalently, when

$n (Γ, p) = 0 for every cycle Γ in Ω and every p \in C ∖ Ω,$

with the index of Integration over a complex chain and the index of a chain.

Remarks

The qualifier is part of the name and is kept in every use. The condition above is about indices, and it is the only notion of simple connectivity defined or used on this page: no notion involving loops, homotopies or a fundamental group is introduced here, and none is invoked in any proof below. Writing "homologically simply connected" everywhere is what keeps that scope visible to a reader who knows the other notions from elsewhere.

$C$ itself is homologically simply connected, because $C ∖ C$ is empty and the condition is then vacuous. More generally the condition constrains a domain only through the points it omits.

Connectedness is part of the definition, since a complex domain is nonempty, open and connected. That is a convenience rather than a necessity for the index condition itself, and every result below that assumes homological simple connectivity therefore has a connected domain available.

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Star-shaped plane domains are homologically simply connected

Statement

Let $U \subseteq C$ be nonempty, open, and star-shaped with respect to some $a \in U$ (Star-shaped open subsets of Euclidean space). Then $U$ is a complex domain and is homologically simply connected (Homologically simply connected complex domains).

In particular every nonempty convex open subset of $C$ is homologically simply connected, since it is star-shaped with respect to each of its points; this covers every open disc and $C$ itself.

Facts & Assumptions

Given: A nonempty open $U \subseteq C$ star-shaped with respect to $a \in U$ ; the plane identification and its segments are those of Complex star-shaped and convex domains are the published Euclidean notions under the identification $C = R^{2}$ .

[L1]

A complex domain is homologically simply connected when every cycle with trace in it is null-homologous in it (Homologically simply connected complex domains), and a cycle $Γ$ with trace in $Ω$ is null-homologous in $Ω$ when $n (Γ, p) = 0$ for every $p \in C ∖ Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L2]

If $Γ$ is a cycle whose trace lies in an open $V$ and $F$ is a primitive on $V$ of a continuous $f$ with $F^{'} = f$ continuous, then $\int_{Γ} f d z = 0$ (The integral of a continuous derivative over a cycle is zero).

[L3]

If $U \subseteq C$ is open and star-shaped with respect to $a \in U$ , every holomorphic $f : U \to C$ has the primitive $F (z) = \int_{ℓ_{a z}} f (ζ) d ζ$ (Every holomorphic function on a star-shaped domain has a primitive).

[L4]

A nonempty open $U \subseteq R^{n}$ is star-shaped with respect to $a \in U$ when $a + t (x - a) \in U$ for every $x \in U$ and $0 \leq t \leq 1$ ; every convex open set is star-shaped with respect to each of its points (Star-shaped open subsets of Euclidean space, A convex subset of $R^{m}$ contains every line segment between two of its points).

[L5]

Constants and the identity are complex differentiable, and linear combinations, products and nonvanishing quotients of functions complex differentiable at a point are complex differentiable there (Linearity, product, reciprocal, and quotient rules for complex derivatives); a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L6]

$n (Γ, p) = (2 π i)^{- 1} \int_{Γ} d z / (z - p)$ for a chain $Γ$ and $p \notin Γ^{*}$ (Integration over a complex chain and the index of a chain), a chain being a finite list of integer-weighted contours (Complex chains, their traces, and cycles).

[L7]

A complex domain is a nonempty, connected, open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ).

[L8]

A subset is path-connected when any two of its points are joined by a continuous map from $[0, 1]$ with image inside it (Paths, path-connected spaces and path components), and a path-connected subset is connected (Every path-connected space is connected, and every path component lies inside a component); a composite of continuous maps is continuous and a function continuous on each member of a finite closed cover is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L9]

A primitive of $f$ on $V$ is a holomorphic $F$ with $F^{'} = f$ on $V$ (A primitive of a complex function on an open set).

Proof

technique · direct

1.1givenL4L7L8

For $x, y \in U$ the maps $t \mapsto x + 2 t (a - x)$ on $[0, \frac{1}{2}]$ and $t \mapsto a + (2 t - 1) (y - a)$ on $[\frac{1}{2}, 1]$ are continuous, take values in $U$ by [L4], and agree at $t = \frac{1}{2}$ with the value $a$ ; the first begins at $x$ and the second ends at $y$ . Thus [L8] joins $x$ to $y$ inside $U$ , making $U$ path-connected, hence connected. With $U$ nonempty and open, [L7] makes it a complex domain.

1.2givenL5L6

Let $Γ$ be a cycle with trace in $U$ and let $p \in C ∖ U$ . Then $z \mapsto 1 / (z - p)$ is holomorphic on $U$ by [L5], since $z - p \neq 0$ there.

2.1step 1.2L3L5L9

By [L3] the function $F (z) = \int_{ℓ_{a z}} \frac{d ζ}{ζ - p}$ is a primitive on $U$ of $z \mapsto 1 / (z - p)$ , and $F^{'}$ equals that function, which is continuous by [L5].

3.1step 1.2step 2.1L2L6

The trace of $Γ$ lies in the open set $U$ , so [L2] applied with $V = U$ , $f (z) = 1 / (z - p)$ and $F$ of step 2.1 gives $\int_{Γ} \frac{d z}{z - p} = 0$ , whence $n (Γ, p) = 0$ by [L6].

4.1step 1.1step 3.1L1L4∎

Since $Γ$ and $p \notin U$ were arbitrary, step 3.1 makes every cycle in $U$ null-homologous in $U$ , so the complex domain of step 1.1 is homologically simply connected by [L1]. A nonempty convex open set is star-shaped with respect to each of its points by [L4], so the same conclusion applies to it.

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On a convex open set the difference quotient is an average of the derivative along the segment

Statement

Let $V \subseteq C$ be open and convex (A convex subset of $R^{m}$ contains every line segment between two of its points) and let $f : V \to C$ be holomorphic. Then for all $z, w \in V$

$f (w) - f (z) = (w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t,$

the integral being the componentwise integral of a continuous $R^{2}$ -valued function of $t$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). In particular, for $w \neq z$ ,

$\frac{f (w) - f (z)}{w - z} = \int_{0}^{1} f^{'} (z + t (w - z)) d t,$

while for $w = z$ the displayed integral equals $f^{'} (z)$ and both sides of the first identity are $0$ .

Facts & Assumptions

Given: An open convex $V \subseteq C$ , a holomorphic $f : V \to C$ and points $z, w \in V$ ; segments in the plane are those of Complex star-shaped and convex domains are the published Euclidean notions under the identification $C = R^{2}$ .

[L1]

[L2]

For a piecewise- $C^{1}$ contour $γ$ and $f$ continuous on its trace, $\int_{γ} f (z) d z = \sum_{j} \int_{t_{j}}^{t_{j + 1}} f (γ (t)) γ_{j}^{'} (t) d t$ (For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals).

[L3]

A holomorphic $f = u + i v$ on an open subset of $C$ has $(u, v)$ of class $C^{k}$ for every natural $k$ , hence smooth (Holomorphic functions are real analytic and smooth in their two real coordinates), and every holomorphic function has complex derivatives of every natural order (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L4]

A subset $U \subseteq R^{m}$ is convex when $(1 - t) x + t y \in U$ for all $x, y \in U$ and $t \in [0, 1]$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L5]

Integrals of $R^{m}$ -valued functions are taken componentwise and are real-linear (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); the real integral is linear in the integrand (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[L6]

A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L7]

A continuous path differentiable with a continuous derivative on each piece of a partition is rectifiable (A continuous piecewise- $C^{1}$ path is rectifiable and its length is the sum of the speed integrals over its pieces).

Proof

technique · direct

1.1givenL1L3L6

By [L3] the derivative $f^{'}$ is again holomorphic on $V$ , hence continuous there by [L6], so $f$ is a primitive of the continuous $f^{'}$ with continuous $F^{'}$ in the sense of [L1].

1.2givenL4L7

The map $ℓ (t) = z + t (w - z)$ on $[0, 1]$ has values in $V$ by [L4], since $V$ is convex, and is differentiable with the constant continuous derivative $w - z$ , so it is a piecewise- $C^{1}$ , hence rectifiable, contour with trace in $V$ ([L7]); its endpoints are $ℓ (0) = z$ and $ℓ (1) = w$ .

2.1step 1.1step 1.2L1

By [L1] applied to $ℓ$ , $\int_{ℓ} f^{'} (ζ) d ζ = f (ℓ (1)) - f (ℓ (0)) = f (w) - f (z)$ .

2.2step 1.1step 1.2L2L5algebra

By [L2] applied to $ℓ$ , whose derivative is the constant $w - z$ , $\int_{ℓ} f^{'} (ζ) d ζ = \int_{0}^{1} f^{'} (z + t (w - z)) (w - z) d t$ , and pulling the complex constant $w - z$ out of the componentwise integral is real linearity, so this equals $(w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t$ .

3.1step 2.1step 2.2L5algebra∎

Steps 2.1 and 2.2 give $f (w) - f (z) = (w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t$ ; dividing by $w - z$ when $w \neq z$ gives the difference-quotient form, and when $w = z$ the integrand is the constant $f^{'} (z)$ , whose integral over $[0, 1]$ is $f^{'} (z)$ by [L5], while both sides of the first identity are $0$ .

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The filled difference quotient of a holomorphic function is jointly continuous

Statement

Let $Ω \subseteq C$ be open and let $f : Ω \to C$ be holomorphic. Define the filled difference quotient $g : Ω \times Ω \to C$ by

$g (ζ, z) = {\begin{cases} \frac{f (ζ) - f (z)}{ζ - z}, & ζ \neq z, \\ f^{'} (z), & ζ = z . \end{cases}$

Then $g$ is continuous on $Ω \times Ω$ , the product carrying the Euclidean metric of $R^{4}$ under the coordinate identification of the plane. Moreover $g (ζ, z) = g (z, ζ)$ for all $ζ, z \in Ω$ .

Facts & Assumptions

Given: An open $Ω \subseteq C$ and a holomorphic $f : Ω \to C$ ; products of subsets of $C$ are read in $R^{4}$ through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

For an open convex $V \subseteq C$ , a holomorphic $f$ on $V$ and $z, w \in V$ , $f (w) - f (z) = (w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t$ , and the displayed integral equals $f^{'} (z)$ when $w = z$ (On a convex open set the difference quotient is an average of the derivative along the segment).

[L2]

With $U$ open, $f$ holomorphic on $U$ and $z \in U$ fixed, the function equal to $(f (ζ) - f (z)) / (ζ - z)$ for $ζ \neq z$ and to $f^{'} (z)$ at $ζ = z$ is continuous on $U$ and holomorphic on $U ∖ {z}$ (The filled difference quotient is continuous at its exceptional point and holomorphic away from it).

[L3]

A holomorphic function is smooth in the real coordinates (Holomorphic functions are real analytic and smooth in their two real coordinates) and has complex derivatives of every natural order (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle), and a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L4]

For an integrable $f : [a, b] \to R^{m}$ with $a \leq b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ (For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable); integrals of vector-valued functions are componentwise and real-linear (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L5]

If $a < b$ , $h \leq H$ pointwise on $[a, b]$ , and both are integrable, then $\int_{a}^{b} h \leq \int_{a}^{b} H$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L6]

A map into $R^{m}$ from a subset of a metric space is continuous at $a$ exactly when the usual $ε$ – $δ$ condition holds with the Euclidean norm (Vector-valued functions $f : A \to R^{m}$ , their limits and continuity, with the dictionary to the metric notions).

[L7]

Sums, products and quotients with nonvanishing denominator of continuous real-valued maps on a topological space are continuous (Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined).

[L8]

$B (x, ρ) = {y : d (x, y) < ρ}$ (Open ball, closed ball and sphere in a metric space), a set is open exactly when each of its points admits a ball inside it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and a set is convex when it contains the segment between any two of its points (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L9]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive); for $z = a + b i$ with $a, b$ real, $Re z = a$ , $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

Proof

technique · direct

1.1givenalgebra

Interchanging $ζ$ and $z$ leaves the off-diagonal formula unchanged, both numerator and denominator changing sign, and leaves the diagonal value $f^{'} (z)$ unchanged; so $g$ is symmetric.

1.2L6L7L9

A complex-valued map on a topological space is continuous exactly when its real and imaginary parts are, by [L6] and [L9]; the real and imaginary parts of a sum, a product and a quotient with nonvanishing denominator of complex-valued maps are the corresponding real polynomial expressions in the parts, with denominator $∣ \cdot ∣^{2}$ , so [L7] makes such combinations of continuous complex-valued maps continuous.

1.3givenL3L6L8choose

Fix $a \in Ω$ and let $ε > 0$ . By [L8] choose $ρ > 0$ with $B (a, ρ) \subseteq Ω$ ; by [L3] the derivative $f^{'}$ is holomorphic, hence continuous, on $Ω$ , so by [L6] there is $ρ^{'}$ with $0 < ρ^{'} \leq ρ$ and $∣ f^{'} (ξ) - f^{'} (a) ∣ \leq ε$ for every $ξ \in B (a, ρ^{'})$ .

2.1givenstep 1.2L2L3L8

On the set $W = {(ζ, z) \in Ω \times Ω : ζ \neq z}$ , which is open by [L8], the maps $(ζ, z) \mapsto f (ζ) - f (z)$ and $(ζ, z) \mapsto ζ - z$ are continuous by [L3] and step 1.2, and [L2] already gives continuity of the filled difference quotient in each variable when the other is fixed; since the second map is nowhere zero on $W$ , step 1.2 makes $g$ continuous on $W$ .

2.2step 1.3L1L8L9

Let $ζ, z \in B (a, ρ^{'})$ . The ball $B (a, ρ^{'})$ is convex by [L8] and [L9] and $f$ is holomorphic on it, so [L1] gives $g (ζ, z) = \int_{0}^{1} f^{'} (z + t (ζ - z)) d t$ both off and on the diagonal, and every point $z + t (ζ - z)$ lies in $B (a, ρ^{'})$ by convexity.

3.1step 2.1step 1.3step 2.2L4L5L6L9∎

Subtracting the constant $g (a, a) = f^{'} (a)$ inside the integral of step 2.2 and applying [L4] and [L5] with the bound of step 1.3 gives $∣ g (ζ, z) - g (a, a) ∣ \leq \int_{0}^{1} ∣ f^{'} (z + t (ζ - z)) - f^{'} (a) ∣ d t \leq ε$ for all $ζ, z \in B (a, ρ^{'})$ ; by [L6] and [L9] this is continuity of $g$ at $(a, a)$ , and with step 2.1 it makes $g$ continuous on all of $Ω \times Ω$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The filled difference quotient is holomorphic in each variable separately

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic and let $g$ be its filled difference quotient (The filled difference quotient of a holomorphic function is jointly continuous). Then for each fixed $z \in Ω$ the map $ζ \mapsto g (ζ, z)$ is holomorphic on the whole of $Ω$ , the point $ζ = z$ included; and by symmetry, for each fixed $ζ \in Ω$ the map $z \mapsto g (ζ, z)$ is holomorphic on $Ω$ .

Facts & Assumptions

Given: An open $Ω \subseteq C$ , a holomorphic $f : Ω \to C$ and its filled difference quotient $g$ .

[L1]

With $U$ open, $f$ holomorphic on $U$ and $z \in U$ fixed, the function equal to $(f (ζ) - f (z)) / (ζ - z)$ for $ζ \neq z$ and to $f^{'} (z)$ at $ζ = z$ is continuous on $U$ and holomorphic on $U ∖ {z}$ ; no holomorphy at the filled point is asserted (The filled difference quotient is continuous at its exceptional point and holomorphic away from it).

[L2]

If $U \subseteq C$ is open, $p \in U$ and $h : U \to C$ is continuous on $U$ and holomorphic on $U ∖ {p}$ , then $h$ is holomorphic on $U$ (A continuous function holomorphic off a single point is holomorphic).

[L3]

The filled difference quotient $g$ of a holomorphic $f$ on $Ω$ is $(f (ζ) - f (z)) / (ζ - z)$ off the diagonal and $f^{'} (z)$ on it, and it satisfies $g (ζ, z) = g (z, ζ)$ (The filled difference quotient of a holomorphic function is jointly continuous).

Proof

technique · direct

1.1givenL1L3

Fix $z \in Ω$ . By [L3] the map $ζ \mapsto g (ζ, z)$ is exactly the function of [L1] for that $z$ , so it is continuous on $Ω$ and holomorphic on $Ω ∖ {z}$ .

2.1step 1.1L2

Applying [L2] with $U = Ω$ , $p = z$ and $h = g (\cdot, z)$ , step 1.1 upgrades that function to a holomorphic function on all of $Ω$ .

3.1step 2.1L3∎

By the symmetry $g (ζ, z) = g (z, ζ)$ of [L3], the map $z \mapsto g (ζ, z)$ for fixed $ζ$ is the map of step 2.1 with the roles of the two arguments exchanged, hence holomorphic on $Ω$ as well.

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The Cauchy transform of a cycle is holomorphic off its trace, with the expected derivatives

Statement

Let $Γ = \sum_{k < r} m_{k} γ_{k}$ be a complex chain with trace $Γ^{*}$ and let $φ$ be continuous on $Γ^{*}$ . Put $V = C ∖ Γ^{*}$ , which is open, and for every natural $n \geq 1$ define the Cauchy transform

$F_{n} (z) = \frac{1}{2 π i} \int_{Γ} \frac{φ (ζ)}{(ζ - z)^{n}} d ζ (z \in V) .$

Then each $F_{n}$ is holomorphic on $V$ and

$F_{n}^{'} (z) = n F_{n + 1} (z) (z \in V) .$

Facts & Assumptions

Given: A complex chain $Γ = \sum_{k < r} m_{k} γ_{k}$ and a continuous $φ$ on its trace.

[L1]

Let $γ : [α, β] \to C$ be a rectifiable contour, let $φ$ be continuous on its trace and let $W \subseteq C$ be open and disjoint from that trace. For every natural $n \geq 1$ the function $z \mapsto (2 π i)^{- 1} \int_{γ} φ (ζ) (ζ - z)^{- n} d ζ$ is holomorphic on $W$ and its derivative is $n$ times the corresponding function with exponent $n + 1$ (Cauchy-kernel contour integrals may be differentiated by a direct difference-quotient estimate).

[L2]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ (Integration over a complex chain and the index of a chain).

[L3]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ of integers and complex contours, and its trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ (Complex chains, their traces, and cycles).

[L4]

Finite linear combinations and products of complex-differentiable functions are complex differentiable, as are reciprocals and quotients wherever their denominators do not vanish; constants and the identity are complex differentiable (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L5]

A compact subset is closed and bounded (A compact subset of a metric space is closed and bounded); the continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); a finite union of compact subsets is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact); a closed bounded interval is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L6]

Negative integer powers are defined exactly for nonzero complex bases (Integer powers in the complex field).

[L7]

A set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L8]

A sum over a finite index set in the additive commutative monoid of $C$ is well posed and additive, with empty sum $0$ ; complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L9]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1givenL3L5L7

Each $γ_{k}^{*}$ is the continuous image of a compact interval, hence compact by [L5], so the trace $Γ^{*}$ of [L3] is a finite union of compact sets, compact by [L5] and closed by [L5]; therefore $V = C ∖ Γ^{*}$ is open by [L7].

1.2givenL2L4L6L9

For $z \in V$ and $ζ \in Γ^{*}$ one has $ζ - z \neq 0$ , so the powers $(ζ - z)^{- n}$ are defined by [L6]. For fixed $z$ , the map $ζ \mapsto (ζ - z)^{- n}$ is holomorphic on $C ∖ {z}$ by repeated products and nonvanishing quotients, using [L4], hence continuous by [L9]; multiplying by the continuous function $φ$ makes the integrand of each $F_{n}$ continuous on $Γ^{*}$ , so [L2] defines $F_{n} (z)$ .

2.1step 1.1step 1.2L1L3

Fix $k < r$ with $m_{k} \neq 0$ . Then $γ_{k}^{*} \subseteq Γ^{*}$ by [L3], so the open set $V$ of step 1.1 is disjoint from $γ_{k}^{*}$ , and $φ$ is continuous on $γ_{k}^{*}$ ; hence [L1] makes $F_{n}^{(k)} (z) = (2 π i)^{- 1} \int_{γ_{k}} φ (ζ) (ζ - z)^{- n} d ζ$ holomorphic on $V$ with $(F_{n}^{(k)})^{'} = n F_{n + 1}^{(k)}$ .

3.1step 2.1L2L4L8∎

By [L2] and [L8], $F_{n} = \sum_{k < r, m_{k} \neq 0} m_{k} F_{n}^{(k)}$ on $V$ , a finite linear combination with constant coefficients of the functions of step 2.1; so [L4] makes $F_{n}$ holomorphic on $V$ with $F_{n}^{'} = \sum_{k} m_{k} (F_{n}^{(k)})^{'} = n \sum_{k} m_{k} F_{n + 1}^{(k)} = n F_{n + 1}$ . The empty chain, and a chain with all coefficients zero, give $F_{n} \equiv 0$ and the identity holds trivially.

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Dixon's glued function is entire and vanishes at infinity

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic and let $Γ$ be a complex chain which is a cycle, with trace in $Ω$ and null-homologous in $Ω$ (Null-homologous cycles and homologous cycles in an open set). Let $g$ be the filled difference quotient of $f$ on $Ω \times Ω$ (The filled difference quotient of a holomorphic function is jointly continuous) and put

$h_{0} (z) = \frac{1}{2 π i} \int_{Γ} g (ζ, z) d ζ (z \in Ω), Ω_{0} = {z \in C ∖ Γ^{*} : n (Γ, z) = 0}, h_{1} (z) = \frac{1}{2 π i} \int_{Γ} \frac{f (ζ)}{ζ - z} d ζ (z \in Ω_{0}) .$

Then $Ω_{0}$ is open, $Ω \cup Ω_{0} = C$ , $h_{0}$ is holomorphic on $Ω$ , $h_{1}$ is holomorphic on $Ω_{0}$ , and $h_{0} = h_{1}$ on $Ω \cap Ω_{0}$ . Consequently the function $h$ equal to $h_{0}$ on $Ω$ and to $h_{1}$ on $Ω_{0}$ is a well-defined entire function; it is bounded, and for every $ε > 0$ there is $R > 0$ with $∣ h (z) ∣ < ε$ whenever $∣ z ∣ > R$ .

Facts & Assumptions

Given: An open $Ω$ , a holomorphic $f : Ω \to C$ , and a cycle $Γ = \sum_{k < r} m_{k} γ_{k}$ with $Γ^{*} \subseteq Ω$ which is null-homologous in $Ω$ ; the plane carries the Euclidean metric of $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

If $γ$ is a rectifiable contour, $W$ is open and $φ$ is continuous on $γ^{*} \times W$ with $φ (w, \cdot)$ holomorphic on $W$ for each $w \in γ^{*}$ , then $z \mapsto \int_{γ} φ (ζ, z) d ζ$ is holomorphic on $W$ (A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic).

[L2]

The filled difference quotient $g$ of a holomorphic $f$ on $Ω$ equals $(f (ζ) - f (z)) / (ζ - z)$ off the diagonal and $f^{'} (z)$ on it, and is continuous on $Ω \times Ω$ (The filled difference quotient of a holomorphic function is jointly continuous).

[L3]

For each fixed $z \in Ω$ the map $ζ \mapsto g (ζ, z)$ is holomorphic on $Ω$ , and for each fixed $ζ \in Ω$ the map $z \mapsto g (ζ, z)$ is holomorphic on $Ω$ (The filled difference quotient is holomorphic in each variable separately).

[L4]

For a complex chain $Γ$ and $φ$ continuous on $Γ^{*}$ , the function $z \mapsto (2 π i)^{- 1} \int_{Γ} φ (ζ) (ζ - z)^{- 1} d ζ$ is holomorphic on $C ∖ Γ^{*}$ (The Cauchy transform of a cycle is holomorphic off its trace, with the expected derivatives).

[L5]

For a cycle $Γ$ the trace $Γ^{*}$ is compact, the index is locally constant on $C ∖ Γ^{*}$ , the set $Ω_{0}$ of points off the trace where the index vanishes is open, and there is $R_{1} > 0$ with $n (Γ, p) = 0$ whenever $∣ p ∣ > R_{1}$ (The index of a cycle is locally constant off its trace and vanishes far from it).

[L6]

A cycle $Γ$ with trace in $Ω$ is null-homologous in $Ω$ when $n (Γ, p) = 0$ for every $p \in C ∖ Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L7]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ , and $n (Γ, p) = (2 π i)^{- 1} \int_{Γ} d z / (z - p)$ for $p \notin Γ^{*}$ (Integration over a complex chain and the index of a chain); a chain is a finite list of integer-weighted contours whose trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ (Complex chains, their traces, and cycles).

[L8]

If $∣ f (z) ∣ \leq M$ on the trace of a rectifiable contour $γ$ , with $M \geq 0$ , then $∣ \int_{γ} f d z ∣ \leq M L (γ)$ (ML estimate: a contour integral is bounded by a supremum bound times path length); complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand).

[L9]

A compact subset is closed and bounded (A compact subset of a metric space is closed and bounded); a continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); a finite union of compact subsets is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact); a closed bounded interval and a closed disc are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L10]

A function holomorphic on all of $C$ is entire, and holomorphy on an open set is complex differentiability at each of its points (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L11]

A set is closed exactly when its complement is open, and a set is open exactly when each of its points admits a ball inside it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space); a set is bounded when it is empty or lies inside a ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L12]

[L13]

[L14]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L15]

Finite linear combinations of holomorphic functions are holomorphic (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Proof

technique · direct

1.1givenL5L9L11

By [L5] the trace $Γ^{*}$ is compact, the set $Ω_{0}$ is open, and there is $R_{1} > 0$ with $n (Γ, p) = 0$ for $∣ p ∣ > R_{1}$ ; by [L9] and [L11] there is also $R_{0} > 0$ with $Γ^{*} \subseteq {w : ∣ w ∣ \leq R_{0}}$ .

1.2givenL6L7

If $p \notin Ω$ then $p \notin Γ^{*}$ , because $Γ^{*} \subseteq Ω$ , and $n (Γ, p) = 0$ by [L6]; so $p \in Ω_{0}$ . Hence $Ω \cup Ω_{0} = C$ .

1.3givenL1L2L3L7L13L15

For each $k < r$ with $m_{k} \neq 0$ the trace $γ_{k}^{*}$ lies in $Γ^{*} \subseteq Ω$ , and $g$ is continuous on $γ_{k}^{*} \times Ω$ by [L2] with $z \mapsto g (w, z)$ holomorphic on $Ω$ for each fixed $w$ by [L3]; so [L1] makes $z \mapsto \int_{γ_{k}} g (ζ, z) d ζ$ holomorphic on $Ω$ , and [L15] therefore makes the finite linear combination $h_{0}$ holomorphic on $Ω$ .

2.1step 1.1L4L14

The restriction of $f$ to $Γ^{*}$ is continuous by [L14], so [L4] makes $z \mapsto (2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- 1} d ζ$ holomorphic on $C ∖ Γ^{*}$ ; since $Ω_{0} \subseteq C ∖ Γ^{*}$ is open by step 1.1, $h_{1}$ is holomorphic on $Ω_{0}$ .

2.2step 1.1L2L7L8L13

Let $z \in Ω \cap Ω_{0}$ . Then $z \notin Γ^{*}$ , so $ζ \neq z$ for every $ζ \in Γ^{*}$ and [L2] gives $g (ζ, z) = (f (ζ) - f (z)) / (ζ - z)$ there; splitting the integral by [L8] and [L13] gives $h_{0} (z) = h_{1} (z) - f (z) n (Γ, z)$ through [L7], and $n (Γ, z) = 0$ because $z \in Ω_{0}$ , so $h_{0} (z) = h_{1} (z)$ .

2.3step 1.1L7L8L9L12L13L14

Let $M_{f} = 0$ when $Γ^{*} = \emptyset$ , and otherwise choose a real $M_{f} \geq 0$ with $∣ f (w) ∣ \leq M_{f}$ for every $w \in Γ^{*}$ ; such a bound exists because $f$ is continuous on the compact trace by [L14], so its image is compact and therefore bounded by [L9]. Put $M (Γ) = \sum_{k < r, m_{k} \neq 0} ∣ m_{k} ∣ L (γ_{k})$ . For $z \in Ω_{0}$ with $∣ z ∣ > R_{0}$ and $ζ \in Γ^{*}$ , [L12] gives $∣ ζ - z ∣ \geq ∣ z ∣ - R_{0} > 0$ , so [L7], [L8] and [L13] give $∣ h_{1} (z) ∣ \leq M_{f} M (Γ) / (2 π (∣ z ∣ - R_{0}))$ .

3.1step 1.2step 1.3step 2.1step 2.2L10

By steps 1.2, 1.3, 2.1 and 2.2 the assignment $h = h_{0}$ on $Ω$ and $h = h_{1}$ on $Ω_{0}$ is a well-defined function on $C$ , and it is complex differentiable at every point because each point lies in one of the two open sets on which the corresponding piece is holomorphic; so $h$ is entire by [L10].

4.1step 3.1step 2.3L7L9L11L14∎

Let $ε > 0$ and take $R = \max {R_{0}, R_{1}} + M_{f} M (Γ) / (2 π ε) + 1$ . For $∣ z ∣ > R$ step 1.1 puts $z$ in $Ω_{0}$ , so $h (z) = h_{1} (z)$ by step 3.1 and step 2.3 gives $∣ h (z) ∣ < ε$ . Taking $ε = 1$ produces one such $R$ , and $h$ is continuous on the compact disc ${∣ z ∣ \leq R}$ by [L9] and [L14], hence bounded there by [L9]; so $h$ is bounded on $C$ . If $Γ^{*} = \emptyset$ then every integral over $Γ$ is $0$ by [L7] and $h$ is identically $0$ , which satisfies both conclusions.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Cauchy's integral formula for a null-homologous cycle

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic, and let $Γ$ be a complex chain which is a cycle, with trace in $Ω$ and null-homologous in $Ω$ . Then for every $z \in Ω ∖ Γ^{*}$

$n (Γ, z) f (z) = \frac{1}{2 π i} \int_{Γ} \frac{f (ζ)}{ζ - z} d ζ .$

No connectedness of $Ω$ is assumed.

Facts & Assumptions

Given: An open $Ω$ , a holomorphic $f : Ω \to C$ , and a cycle $Γ$ with $Γ^{*} \subseteq Ω$ which is null-homologous in $Ω$ .

[L1]

With $g$ the filled difference quotient of $f$ , the function $h$ equal to $(2 π i)^{- 1} \int_{Γ} g (ζ, z) d ζ$ on $Ω$ and to $(2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- 1} d ζ$ on $Ω_{0} = {z \notin Γ^{*} : n (Γ, z) = 0}$ is a well-defined entire function; it is bounded, and for every $ε > 0$ there is $R > 0$ with $∣ h (z) ∣ < ε$ whenever $∣ z ∣ > R$ (Dixon's glued function is entire and vanishes at infinity).

[L2]

Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

[L3]

If $f$ is continuous on the trace of a complex chain, then $\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ ; and for $z \notin Γ^{*}$ one has $n (Γ, z) = (2 π i)^{- 1} \int_{Γ} d ζ / (ζ - z)$ (Integration over a complex chain and the index of a chain). A chain is a finite list of integer-weighted contours with trace the union of the $γ_{k}^{*}$ having $m_{k} \neq 0$ (Complex chains, their traces, and cycles).

[L4]

A cycle $Γ$ with trace in $Ω$ is null-homologous in $Ω$ when $n (Γ, p) = 0$ for every $p \in C ∖ Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L5]

Complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand); finite sums in the additive commutative monoid of $C$ are additive, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L6]

The filled difference quotient $g$ of a holomorphic $f$ on $Ω$ equals $(f (ζ) - f (z)) / (ζ - z)$ off the diagonal and $f^{'} (z)$ on it (The filled difference quotient of a holomorphic function is jointly continuous).

[L7]

A function holomorphic on all of $C$ is entire (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L8]

A holomorphic function is continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1givenL1L2L7

By [L1] the glued function $h$ is entire and bounded, so [L2] makes it a constant $c$ .

1.2givenL1

By [L1], for every $ε > 0$ there is $R > 0$ with $∣ h (z) ∣ < ε$ for $∣ z ∣ > R$ ; such $z$ exist, so $∣ c ∣ < ε$ for every $ε > 0$ and therefore $c = 0$ .

1.3givenL3L5L6L8

Let $z \in Ω ∖ Γ^{*}$ . Since $f$ is holomorphic on $Ω$ , [L8] makes it continuous on $Ω$ , hence on the trace of $Γ$ . Then $ζ \neq z$ for every $ζ \in Γ^{*}$ , so [L6] gives $g (ζ, z) = (f (ζ) - f (z)) / (ζ - z)$ on the trace, and [L3] with [L5] splits the defining integral into $h (z) = (2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- 1} d ζ - f (z) n (Γ, z)$ .

2.1step 1.1step 1.2step 1.3L1L4∎

Steps 1.1, 1.2 and 1.3 give $0 = (2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- 1} d ζ - n (Γ, z) f (z)$ , which is the stated formula; nothing in the argument used connectedness of $Ω$ , and the hypothesis that $Γ$ is null-homologous entered only through [L1] and [L4].

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Cauchy's theorem for a null-homologous cycle

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic, and let $Γ$ be a complex chain which is a cycle, with trace in $Ω$ and null-homologous in $Ω$ . Then

$\int_{Γ} f (z) d z = 0.$

Facts & Assumptions

Given: An open $Ω$ , a holomorphic $f : Ω \to C$ , and a cycle $Γ$ with $Γ^{*} \subseteq Ω$ which is null-homologous in $Ω$ ; the plane is read as $R^{2}$ through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

Under the hypotheses above, $n (Γ, z) f (z) = (2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- 1} d ζ$ for every $z \in Ω ∖ Γ^{*}$ (Cauchy's integral formula for a null-homologous cycle).

[L2]

Products of functions complex differentiable at a point are complex differentiable there, and constants have derivative $0$ (Linearity, product, reciprocal, and quotient rules for complex derivatives); a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L3]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ , and $n (Γ, z) = (2 π i)^{- 1} \int_{Γ} d ζ / (ζ - z)$ for $z \notin Γ^{*}$ (Integration over a complex chain and the index of a chain); a chain is a finite list of integer-weighted complex contours and its trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ (Complex chains, their traces, and cycles).

[L4]

A cycle $Γ$ with trace in $Ω$ is null-homologous in $Ω$ when $n (Γ, p) = 0$ for every $p \in C ∖ Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L5]

Complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand).

[L6]

A compact subset is closed and bounded (A compact subset of a metric space is closed and bounded); a finite union of compact subsets is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact); a continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); a closed bounded interval is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L7]

A topological space is connected exactly when its only clopen subsets are the empty set and the whole space (For a topological space the following agree: no separation exists, the only clopen subsets are $\emptyset$ and $X$ , and every continuous map to the two-point discrete space is constant).

[L8]

For $n \geq 1$ , $R^{n}$ is polygonally connected and connected ( $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected).

[L9]

A set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and a subset is bounded when it is empty or lies inside a ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Proof

technique · direct

1.1givenL3

If $Ω = \emptyset$ then $Γ^{*} = \emptyset$ , so by [L3] every $m_{k}$ is zero or the list is empty and $\int_{Γ} f d z = 0$ ; assume from now on that $Ω \neq \emptyset$ .

1.2givenL3L6

The trace $Γ^{*}$ is a finite union of continuous images of compact intervals by [L3], hence compact by [L6], and therefore closed and bounded by [L6].

2.1step 1.1step 1.2L7L8L9

There is a point $z \in Ω ∖ Γ^{*}$ . Indeed $Γ^{*} \subseteq Ω$ , so $Ω ∖ Γ^{*} = \emptyset$ would force $Ω = Γ^{*}$ ; by step 1.2 that set is closed, and $Ω$ is open, so $Ω$ would be a nonempty clopen subset of $C$ which is bounded by step 1.2 and [L9], hence different from $C$ . That contradicts [L7] and [L8], since $C$ is connected and its only clopen subsets are $\emptyset$ and $C$ . The trace is not asserted to have empty interior anywhere in this argument.

3.1step 2.1L1L2L4

Fix such a $z$ and put $F (ζ) = (ζ - z) f (ζ)$ for $ζ \in Ω$ , which is holomorphic on $Ω$ by [L2] and satisfies $F (z) = 0$ . Since $Γ$ is null-homologous in $Ω$ by the hypothesis and [L4], [L1] applies to $F$ at the point $z$ and gives $0 = n (Γ, z) F (z) = (2 π i)^{- 1} \int_{Γ} F (ζ) (ζ - z)^{- 1} d ζ$ .

4.1step 3.1L3L5∎

On the trace $ζ \neq z$ , so $F (ζ) (ζ - z)^{- 1} = f (ζ)$ there, and the integrand of step 3.1 is $f$ itself; hence $\int_{Γ} f (ζ) d ζ = 0$ by [L3] and [L5].

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Holomorphic integrals agree on homologous cycles

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic, and let $Γ_{0}, Γ_{1}$ be complex chains which are cycles with traces in $Ω$ and which are homologous in $Ω$ (Null-homologous cycles and homologous cycles in an open set). Then

$\int_{Γ_{0}} f (z) d z = \int_{Γ_{1}} f (z) d z .$

Facts & Assumptions

Given: An open $Ω$ , a holomorphic $f : Ω \to C$ , and cycles $Γ_{0}, Γ_{1}$ with traces in $Ω$ , homologous in $Ω$ .

[L1]

If $Γ$ is a cycle with trace in an open $Ω$ , null-homologous in $Ω$ , and $f$ is holomorphic on $Ω$ , then $\int_{Γ} f d z = 0$ (Cauchy's theorem for a null-homologous cycle).

[L2]

Two cycles with traces in $Ω$ are homologous in $Ω$ when their difference is null-homologous in $Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L3]

$(Γ_{1} + Γ_{2})^{*} = Γ_{1}^{*} \cup Γ_{2}^{*}$ and $(- Γ)^{*} = Γ^{*}$ ; the sum of two cycles and the negative of a cycle are cycles; and for $f$ continuous on the traces involved, $\int_{Γ_{1} + Γ_{2}} f d z = \int_{Γ_{1}} f d z + \int_{Γ_{2}} f d z$ and $\int_{- Γ} f d z = - \int_{Γ} f d z$ (Chain integration and the index are additive in the chain, and reverse with it).

[L4]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ (Integration over a complex chain and the index of a chain), a chain being a finite list of integer-weighted complex contours (Complex chains, their traces, and cycles).

[L5]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1givenL3L4L5

By [L3] the chain $Γ_{0} - Γ_{1}$ is a cycle and its trace is $Γ_{0}^{*} \cup Γ_{1}^{*}$ , which lies in $Ω$ ; and $f$ is continuous on that trace by [L5].

1.2givenL2

By [L2] the chain $Γ_{0} - Γ_{1}$ is null-homologous in $Ω$ , since $Γ_{0}$ and $Γ_{1}$ are homologous there.

2.1step 1.1step 1.2L1

Steps 1.1 and 1.2 put $Γ_{0} - Γ_{1}$ under the hypotheses of [L1], so $\int_{Γ_{0} - Γ_{1}} f d z = 0$ .

3.1step 2.1L3∎

By [L3] the left-hand side of step 2.1 equals $\int_{Γ_{0}} f d z - \int_{Γ_{1}} f d z$ , so the two integrals agree.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The higher-derivative form of the global Cauchy formula

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic, and let $Γ$ be a complex chain which is a cycle, with trace in $Ω$ and null-homologous in $Ω$ . Then for every natural number $m$ and every $z \in Ω ∖ Γ^{*}$

$n (Γ, z) f^{(m)} (z) = \frac{m!}{2 π i} \int_{Γ} \frac{f (ζ)}{(ζ - z)^{m + 1}} d ζ,$

with $f^{(0)} = f$ . The case $m = 0$ is the integral formula already proved.

Facts & Assumptions

Given: An open $Ω$ , a holomorphic $f : Ω \to C$ , and a cycle $Γ$ with $Γ^{*} \subseteq Ω$ which is null-homologous in $Ω$ .

[L1]

Under these hypotheses, $n (Γ, z) f (z) = (2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- 1} d ζ$ for every $z \in Ω ∖ Γ^{*}$ (Cauchy's integral formula for a null-homologous cycle).

[L2]

For a chain $Γ$ and $φ$ continuous on $Γ^{*}$ , the functions $F_{j} (z) = (2 π i)^{- 1} \int_{Γ} φ (ζ) (ζ - z)^{- j} d ζ$ are holomorphic on $C ∖ Γ^{*}$ for every natural $j \geq 1$ and satisfy $F_{j}^{'} = j F_{j + 1}$ (The Cauchy transform of a cycle is holomorphic off its trace, with the expected derivatives).

[L3]

For a cycle $Γ$ the trace is compact, the index is constant on every connected component of $C ∖ Γ^{*}$ , and each such component is open (The index of a cycle is locally constant off its trace and vanishes far from it).

[L4]

A holomorphic function is smooth in the real coordinates (Holomorphic functions are real analytic and smooth in their two real coordinates) and has complex derivatives of every natural order (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle); a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L5]

If a property holds at $0$ and passes from $j$ to $j + 1$ , it holds for every natural number (The principle of mathematical induction).

[L6]

$0! = 1$ and $(j + 1)! = j! (j + 1)$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L7]

A constant multiple of a function complex differentiable at a point is complex differentiable there with the corresponding derivative (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L8]

Negative integer powers are defined exactly for nonzero complex bases (Integer powers in the complex field).

[L9]

$n (Γ, z) = (2 π i)^{- 1} \int_{Γ} d ζ / (ζ - z)$ for $z \notin Γ^{*}$ (Integration over a complex chain and the index of a chain), and null-homology in $Ω$ means the index vanishes at every point outside $Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L10]

The connected component of a point is the union of all connected subsets containing it (Connected components, quasicomponents, and totally disconnected spaces), and a set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct

1.1givenL2L3L4L8L10

The trace $Γ^{*}$ is compact by [L3], hence closed, so $C ∖ Γ^{*}$ is open by [L10] and $Ω ∖ Γ^{*}$ is open. The restriction of $f$ to $Γ^{*}$ is continuous by [L4], so the functions $F_{j} (z) = (2 π i)^{- 1} \int_{Γ} f (ζ) (ζ - z)^{- j} d ζ$ of [L2] are defined and holomorphic on $C ∖ Γ^{*}$ with $F_{j}^{'} = j F_{j + 1}$ , the powers being legitimate by [L8].

1.2givenL4

By [L4] the function $f$ has complex derivatives $f^{(m)}$ of every natural order on $Ω$ .

2.1step 1.1L5L6L7

An induction on $j$ ([L5]) using $F_{1}^{'} = F_{2}$ , the relation $F_{j}^{'} = j F_{j + 1}$ of step 1.1, [L6] and [L7] gives $F_{1}^{(j)} = j! F_{j + 1}$ on $C ∖ Γ^{*}$ for every natural $j$ , the case $j = 0$ reading $F_{1} = 0! F_{1}$ .

2.2step 1.1L3L9L10

Fix $z_{0} \in Ω ∖ Γ^{*}$ and let $C$ be the connected component of $z_{0}$ in $C ∖ Γ^{*}$ . By [L3] the set $C$ is open and $n (Γ, \cdot)$ is a constant $k$ on it, so $W = C \cap Ω$ is an open subset of $Ω ∖ Γ^{*}$ containing $z_{0}$ on which the index has the constant value $k$ .

3.1step 1.2step 2.2L1L7

By [L1] the identity $k f = F_{1}$ holds on $W$ ; both sides are holomorphic there by steps 1.1 and 1.2, and complex differentiation is a local operation, so differentiating $m$ times on $W$ and using [L7] gives $k f^{(m)} = F_{1}^{(m)}$ on $W$ .

4.1step 2.1step 3.1L1L6∎

Combining step 3.1 with step 2.1 at the point $z_{0}$ gives $n (Γ, z_{0}) f^{(m)} (z_{0}) = k f^{(m)} (z_{0}) = m! F_{m + 1} (z_{0})$ , which is the displayed formula; since $z_{0} \in Ω ∖ Γ^{*}$ was arbitrary and $m$ was an arbitrary natural number, the formula holds throughout, and at $m = 0$ it is [L1] again by [L6].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every holomorphic function on a homologically simply connected domain has a primitive

Statement

Let $Ω$ be a homologically simply connected complex domain (Homologically simply connected complex domains). Then every holomorphic $f : Ω \to C$ has a primitive on $Ω$ (A primitive of a complex function on an open set): there is a holomorphic $F : Ω \to C$ with $F^{'} = f$ .

Facts & Assumptions

Given: A homologically simply connected complex domain $Ω$ and a holomorphic $f : Ω \to C$ .

[L1]

If $Γ$ is a cycle with trace in an open $Ω$ , null-homologous in $Ω$ , and $f$ is holomorphic on $Ω$ , then $\int_{Γ} f d z = 0$ (Cauchy's theorem for a null-homologous cycle).

[L2]

A complex domain is homologically simply connected when every cycle with trace in it is null-homologous in it (Homologically simply connected complex domains, Null-homologous cycles and homologous cycles in an open set).

[L3]

A list of closed complex contours is a cycle; in particular a single closed contour, taken as the list of length $1$ with coefficient $1$ , is a cycle, and its trace is the trace of that contour (Complex chains, their traces, and cycles).

[L4]

For a chain consisting of the single closed contour $γ$ with coefficient $1$ , $\int_{Γ} f d z = \int_{γ} f d z$ (Integration over a complex chain and the index of a chain).

[L5]

For a complex domain $U$ and a continuous $f : U \to C$ , the following are equivalent: $f$ has a primitive on $U$ ; the integral of $f$ along rectifiable contours in $U$ depends only on the endpoints; the integral of $f$ around every closed rectifiable contour in $U$ is $0$ (For a continuous function on a complex domain, endpoint independence, zero closed-contour integrals, and existence of a primitive are equivalent).

[L6]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L7]

A complex domain is a nonempty, connected, open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ).

Proof

technique · direct

1.1givenL3

Let $γ$ be a closed rectifiable contour with trace in $Ω$ , and let $Γ$ be the chain consisting of $γ$ with coefficient $1$ . By [L3] that chain is a cycle whose trace is $γ^{*} \subseteq Ω$ .

2.1step 1.1L1L2L4

By [L2] the cycle $Γ$ is null-homologous in $Ω$ , so [L1] gives $\int_{Γ} f d z = 0$ , and [L4] rewrites this as $\int_{γ} f d z = 0$ .

3.1givenstep 2.1L5L6L7∎

The set $Ω$ is a complex domain by [L7] and $f$ is continuous on it by [L6], so [L5] applies; step 2.1 supplies its third condition for every closed rectifiable contour in $Ω$ , and the equivalence therefore yields a primitive $F$ of $f$ on $Ω$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm

Statement

Let $Ω$ be a homologically simply connected complex domain and let $f : Ω \to C$ be holomorphic and nowhere zero. Then there is a holomorphic $L : Ω \to C$ with

$\exp (L (z)) = f (z) (z \in Ω),$

and any two such functions differ by a constant lying in $2 π i Z$ .

Facts & Assumptions

Given: A homologically simply connected complex domain $Ω$ and a holomorphic nowhere-zero $f : Ω \to C$ .

[L1]

Every holomorphic function on a homologically simply connected complex domain has a primitive there (Every holomorphic function on a homologically simply connected domain has a primitive), that is a holomorphic $F$ with $F^{'}$ equal to the function (A primitive of a complex function on an open set).

[L2]

The complex exponential maps $C$ onto $C ∖ {0}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

[L3]

The complex exponential is entire with $\exp^{'} = \exp$ (The complex exponential is entire and its complex derivative is itself).

[L4]

If $f$ is complex differentiable at $a$ and $g$ at $f (a)$ , then $(g \circ f)^{'} (a) = g^{'} (f (a)) f^{'} (a)$ (The chain rule for complex derivatives).

[L5]

Linear combinations, products and nonvanishing quotients of functions complex differentiable at a point are complex differentiable there, with the usual formulas (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L6]

If $U$ is a complex domain and $g : U \to C$ is holomorphic with $g^{'} \equiv 0$ , then $g$ is constant on $U$ (A holomorphic function with zero derivative on a domain is constant).

[L7]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L8]

Every holomorphic function has complex derivatives of all natural orders locally (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle); a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L9]

[L10]

A complex domain is a nonempty, connected, open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ), and a homologically simply connected domain is such a domain in which every cycle is null-homologous (Homologically simply connected complex domains).

[L11]

For $z = a + b i$ with $a, b$ real, $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus); the integers form an ordered commutative ring and are discrete in $R$ , so if $m < n$ then $m + \frac{1}{2}$ lies strictly between them and is not an integer (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[L12]

For real $x, y$ , $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L13]

$\exp (z + w) = \exp z \exp w$ for all complex $z, w$ , hence $\exp (- w) \exp (w) = 1$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

Proof

technique · direct

1.1givenL5L8

By [L8] the derivative $f^{'}$ is holomorphic on $Ω$ , and $f$ is nowhere zero, so the logarithmic derivative $f^{'} / f$ is holomorphic on $Ω$ by [L5].

1.2givenL2L10

Fix $z_{0} \in Ω$ , which is nonempty by [L10]. Since $f (z_{0}) \neq 0$ , [L2] gives $w_{0} \in C$ with $\exp (w_{0}) = f (z_{0})$ .

2.1step 1.1step 1.2L1L5

By [L1] the function $f^{'} / f$ of step 1.1 has a primitive $G$ on $Ω$ ; put $F = G - G (z_{0}) + w_{0}$ , so that $F$ is holomorphic with $F^{'} = f^{'} / f$ and $F (z_{0}) = w_{0}$ .

3.1step 2.1L3L4L5L6L10

The function $u = f \exp (- F)$ is holomorphic on $Ω$ by [L3], [L4] and [L5], and $u^{'} = f^{'} \exp (- F) - f F^{'} \exp (- F) = (f^{'} - f \cdot (f^{'} / f)) \exp (- F) = 0$ throughout $Ω$ ; so $u$ is a constant by [L6] and [L10].

4.1step 1.2step 2.1step 3.1L13

Evaluating at $z_{0}$ gives that constant: $u (z_{0}) = f (z_{0}) \exp (- w_{0}) = \exp (w_{0}) \exp (- w_{0}) = 1$ by step 1.2 and [L13], so $f = \exp (F)$ on $Ω$ and $L = F$ has the required property.

5.1step 4.1L7L8L9L10L11L12∎

If $L_{1}, L_{2}$ are holomorphic on $Ω$ with $\exp \circ L_{1} = \exp \circ L_{2} = f$ , then $L_{1} - L_{2}$ takes values in $2 π i Z$ by [L7]; it is continuous by [L8], so $Im (L_{1} - L_{2}) / (2 π)$ is a continuous integer-valued real function by [L11] and [L12], and [L9] with [L10] and [L11] forces it to be constant on the connected $Ω$ . Hence $L_{1} - L_{2}$ is a constant in $2 π i Z$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A nonvanishing holomorphic function on such a domain has holomorphic roots of every positive order

Statement

Let $Ω$ be a homologically simply connected complex domain, let $f : Ω \to C$ be holomorphic and nowhere zero, and let $m$ be a natural number with $m \geq 1$ . Then there is a holomorphic, nowhere-zero $q : Ω \to C$ with

$q (z)^{m} = f (z) (z \in Ω) .$

One such $q$ is $\exp (L / m)$ for any holomorphic logarithm $L$ of $f$ ; replacing $L$ by another holomorphic logarithm of $f$ multiplies $q$ by an $m$ th root of unity.

Facts & Assumptions

Given: A homologically simply connected complex domain $Ω$ , a holomorphic nowhere-zero $f : Ω \to C$ , and a natural $m \geq 1$ .

[L1]

On a homologically simply connected complex domain, a holomorphic nowhere-zero $f$ admits a holomorphic $L$ with $\exp \circ L = f$ , and any two such differ by a constant in $2 π i Z$ (A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm, Homologically simply connected complex domains).

[L2]

$\exp (z + w) = \exp z \exp w$ for all complex $z, w$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[L3]

The complex exponential is entire with $\exp^{'} = \exp$ (The complex exponential is entire and its complex derivative is itself).

[L4]

The composite of functions complex differentiable at the relevant points is complex differentiable, with $(g \circ f)^{'} (a) = g^{'} (f (a)) f^{'} (a)$ (The chain rule for complex derivatives); constant multiples of complex differentiable functions are complex differentiable (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L5]

For a natural $m \geq 1$ , the $m$ th roots of unity are exactly the numbers $\exp (2 π i k / m)$ for natural $k$ with $0 \leq k < m$ (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ).

[L6]

Natural powers satisfy $z^{0} = 1$ and $z^{j + 1} = z^{j} z$ (Integer powers in the complex field).

[L7]

If a property holds at $0$ and passes from $j$ to $j + 1$ , it holds for every natural number (The principle of mathematical induction).

[L8]

For real $x, y$ , $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L9]

$\ker (\exp) = 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

Proof

technique · direct

1.1givenL1L3L4

By [L1] fix a holomorphic $L$ on $Ω$ with $\exp \circ L = f$ , and put $q = \exp (L / m)$ , which is holomorphic on $Ω$ by [L3] and [L4].

1.2L8

The exponential never vanishes, since $∣ \exp v ∣ = e^{Re v} > 0$ by [L8]; so $q$ is nowhere zero.

2.1step 1.1L2L6L7

An induction on $j$ ([L7]) using [L2] and [L6] gives $\exp (v)^{j} = \exp (j v)$ for every complex $v$ and every natural $j$ , the case $j = 0$ reading $1 = \exp (0)$ . Taking $j = m$ and $v = L (z) / m$ gives $q (z)^{m} = \exp (L (z)) = f (z)$ .

3.1step 1.1step 2.1L1L2L5L9∎

If $L_{1}$ is another holomorphic logarithm of $f$ then $L_{1} = L + 2 π i k$ for a fixed integer $k$ by [L1] and [L9], so $\exp (L_{1} / m) = \exp (L / m) \exp (2 π i k / m)$ by [L2], and $\exp (2 π i k / m)$ is an $m$ th root of unity by [L5].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Equivalent characterisations of a homologically simply connected domain

Statement

Let $U$ be a complex domain. The following are equivalent.

$U$ is homologically simply connected: every complex chain which is a cycle with trace in $U$ is null-homologous in $U$ .
Every holomorphic function on $U$ has a primitive on $U$ .
Every holomorphic nowhere-zero function on $U$ has a holomorphic logarithm on $U$ .
For every $p \in C ∖ U$ , the function $z \mapsto 1 / (z - p)$ has a primitive on $U$ .
$\int_{Γ} f (z) d z = 0$ for every holomorphic $f$ on $U$ and every cycle $Γ$ with trace in $U$ .
$\int_{Γ} \frac{d z}{z - p} = 0$ for every cycle $Γ$ with trace in $U$ and every $p \in C ∖ U$ .

Facts & Assumptions

Given: A complex domain $U$ .

[L1]

[L2]

Every holomorphic function on a homologically simply connected complex domain has a primitive there (Every holomorphic function on a homologically simply connected domain has a primitive).

[L3]

[L4]

If $L$ and $h$ are holomorphic on an open set with $\exp \circ L = h$ , then $h$ is nowhere zero and $L^{'} = h^{'} / h$ ; for $h (z) = z - p$ on a set missing $p$ this gives $L^{'} (z) = 1 / (z - p)$ (A holomorphic logarithm is a primitive of the logarithmic derivative).

[L5]

[L6]

A primitive of $f$ on $V$ is a holomorphic $F$ with $F^{'} = f$ on $V$ (A primitive of a complex function on an open set).

[L7]

[L8]

A complex domain is a nonempty, connected, open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ).

[L9]

The complex exponential is entire with $\exp^{'} = \exp$ (The complex exponential is entire and its complex derivative is itself), and if $f : V \to W$ and $g : W \to C$ are complex differentiable at the relevant points, then $(g \circ f)^{'} (a) = g^{'} (f (a)) f^{'} (a)$ (The chain rule for complex derivatives).

[L10]

The complex exponential maps $C$ onto $C ∖ {0}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

[L11]

A holomorphic function with vanishing derivative on a complex domain is constant there (A holomorphic function with zero derivative on a domain is constant).

[L12]

For real $x, y$ , $∣ \exp (x + i y) ∣ = e^{x}$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L13]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there), and every holomorphic function has complex derivatives of all natural orders locally (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L14]

$\exp (z + w) = \exp z \exp w$ , so $\exp (w) \exp (- w) = 1$ for every complex $w$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

Proof

technique · direct

1.1givenL1L2

Condition 1 implies condition 2: this is [L2] applied to the domain $U$ , which condition 1 makes homologically simply connected by [L1].

1.2givenL6L7L8L9L10L11L12L13L14

Condition 2 implies condition 3, argued from condition 2 alone and not from the theorem about homologically simply connected domains. Let $f$ be holomorphic and nowhere zero on $U$ ; by [L13] the derivative $f^{'}$ is holomorphic, so $f^{'} / f$ is holomorphic on $U$ by [L7], and condition 2 supplies a primitive $G$ with $G^{'} = f^{'} / f$ ([L6]). Fix $z_{0} \in U$ , nonempty by [L8], and use [L10] to pick $w_{0}$ with $\exp (w_{0}) = f (z_{0})$ ; put $F = G - G (z_{0}) + w_{0}$ . Since [L12] shows the exponential never vanishes, $\exp (- F)$ is holomorphic and nowhere zero, so $f \exp (- F)$ is holomorphic with derivative $(f^{'} - f (f^{'} / f)) \exp (- F) = 0$ on $U$ by [L7] and [L9], hence constant by [L11] and [L8]; its value at $z_{0}$ is $\exp (w_{0}) \exp (- w_{0}) = 1$ by [L14], so $\exp \circ F = f$ .

1.3givenL4L6L7

Condition 3 implies condition 4. Let $p \in C ∖ U$ ; then $h (z) = z - p$ is holomorphic and nowhere zero on $U$ by [L7], so condition 3 gives a holomorphic $g$ on $U$ with $\exp \circ g = h$ , and [L4] gives $g^{'} (z) = 1 / (z - p)$ ; thus $g$ is a primitive of $z \mapsto 1 / (z - p)$ on $U$ in the sense of [L6].

1.4givenL3L6L7L13

Condition 4 implies condition 6. Let $Γ$ be a cycle with trace in $U$ and $p \in C ∖ U$ . Condition 4 supplies a primitive of $z \mapsto 1 / (z - p)$ on the open set $U$ , whose derivative is that function and is continuous by [L7] and [L13]; so [L3] gives $\int_{Γ} d z / (z - p) = 0$ .

1.5givenL1L5

Condition 6 implies condition 1. For a cycle $Γ$ with trace in $U$ and $p \in C ∖ U$ , condition 6 and [L5] give $n (Γ, p) = 0$ ; by [L1] that is exactly null-homology of $Γ$ in $U$ , for every such $Γ$ , which is condition 1.

1.6givenL3L6L13

Condition 2 implies condition 5. Given a holomorphic $f$ on $U$ and a cycle $Γ$ with trace in $U$ , condition 2 supplies a primitive $F$ with $F^{'} = f$ , continuous by [L13]; so [L3] gives $\int_{Γ} f d z = 0$ .

1.7givenL5L7

Condition 5 implies condition 6. For $p \in C ∖ U$ the function $z \mapsto 1 / (z - p)$ is holomorphic on $U$ by [L7], so condition 5 applied to it gives $\int_{Γ} d z / (z - p) = 0$ for every cycle $Γ$ with trace in $U$ .

2.1step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7∎

Steps 1.1, 1.2, 1.3, 1.4 and 1.5 close the cycle of implications $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 6 \Rightarrow 1$ , so conditions 1, 2, 3, 4 and 6 are equivalent; steps 1.6 and 1.7 insert condition 5 between conditions 2 and 6, which are already known equivalent, so all six conditions are equivalent.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The principal logarithm is the normalised holomorphic branch on the slit plane

Statement

Let $S = C ∖ {x \in R : x \leq 0}$ be the slit plane. Then $S$ is a complex domain, star-shaped with respect to $1$ , and homologically simply connected. The principal logarithm $Log$ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers) is the unique holomorphic $F : S \to C$ with

$\exp (F (z)) = z (z \in S) and F (1) = 0,$

and it satisfies ${Log}^{'} (z) = 1 / z$ on $S$ .

Facts & Assumptions

Given: The slit plane $S = C ∖ {x \in R : x \leq 0}$ ; segments and star-shapedness in the plane are those of Complex star-shaped and convex domains are the published Euclidean notions under the identification $C = R^{2}$ .

[L1]

[L2]

A nonempty open star-shaped subset of $C$ is a complex domain and is homologically simply connected (Star-shaped plane domains are homologically simply connected).

[L3]

If $L$ and $h$ are holomorphic on an open set with $\exp \circ L = h$ , then $h$ is nowhere zero and $L^{'} = h^{'} / h$ (A holomorphic logarithm is a primitive of the logarithmic derivative).

[L4]

For $z \neq 0$ with principal polar form $z = r (\cos θ + i \sin θ)$ and $- π < θ \leq π$ , $Log z = \log r + i θ$ (Complex logarithms, the principal logarithm, and principal and multivalued complex powers).

[L5]

Every $z \neq 0$ has a unique representation $z = r (\cos θ + i \sin θ)$ with $r = ∣ z ∣ > 0$ and $- π < θ \leq π$ (Every nonzero complex number has a unique polar form $r (\cos θ + i \sin θ)$ with $r > 0$ and $- π < θ \leq π$ ).

[L6]

For $z \neq 0$ the solutions of $\exp w = z$ are exactly $Log z + 2 π i k$ for $k \in Z$ (All logarithms of $z \neq 0$ are $Log z + 2 π i k$ , $k \in Z$ ).

[L7]

[L8]

A continuous real function on $[a, b]$ attains every value between $f (a)$ and $f (b)$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[L9]

A nonempty open $U \subseteq R^{n}$ is star-shaped with respect to $a \in U$ when $a + t (x - a) \in U$ for every $x \in U$ and $0 \leq t \leq 1$ (Star-shaped open subsets of Euclidean space).

[L10]

A complex domain is a nonempty, connected, open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ).

[L11]

For $x > 0$ , $\log x$ is the unique real $y$ with $\exp y = x$ (The natural logarithm as the inverse of the exponential function).

[L12]

$\ker (\exp) = 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L13]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L14]

For $z = a + b i$ with $a, b$ real, $Re z = a$ , $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

[L15]

A set is open exactly when each of its points admits a ball inside it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[L16]

$Z$ is the ring of integers (The integers as equivalence classes of pairs of naturals).

Proof

technique · direct

1.1givenL14L15

$S$ is open: if $z = x + i y \in S$ with $y \neq 0$ then the ball of radius $∣ y ∣$ about $z$ contains no real number, and if $y = 0$ then $x > 0$ and the ball of radius $x$ about $z$ contains no real number $\leq 0$ ; in both cases [L14] and [L15] put a ball around $z$ inside $S$ . It is nonempty, since $1 \in S$ .

1.2givenL9L14

$S$ is star-shaped with respect to $1$ in the sense of [L9]: for $z \in S$ and $0 \leq t \leq 1$ put $w = 1 + t (z - 1)$ . If $w$ were a real number $\leq 0$ then $Im w = t Im z = 0$ by [L14]; $t = 0$ gives $w = 1 > 0$ , so $t > 0$ and $Im z = 0$ , making $z$ a real number, necessarily $z > 0$ because $z \in S$ ; but then $w = (1 - t) + t z > 0$ , a contradiction.

2.1step 1.1step 1.2L2L10

By steps 1.1 and 1.2 and [L2], $S$ is a complex domain and is homologically simply connected.

3.1step 2.1L1L12

The identity function $h (z) = z$ is holomorphic and nowhere zero on $S$ , because $0 \notin S$ , so [L1] gives a holomorphic $G$ on $S$ with $\exp \circ G = h$ ; since $\exp (G (1)) = 1$ , [L12] puts $G (1)$ in $2 π i Z$ , and $F : = G - G (1)$ is holomorphic with $\exp \circ F = h$ and $F (1) = 0$ .

4.1step 1.2step 3.1L7L8L11L13L14

Fix $z \in S$ and let $ϕ (t) = Im F (1 + t (z - 1))$ for $t \in [0, 1]$ ; the segment lies in $S$ by step 1.2, and $ϕ$ is continuous by [L13] and [L14], with $ϕ (0) = 0$ . If $∣ ϕ (1) ∣ \geq π$ then [L8] gives $t$ with $ϕ (t) = π$ or $ϕ (t) = - π$ ; writing $w = 1 + t (z - 1)$ and using $\exp (F (w)) = w$ together with [L7], [L11] and [L14] gives $w = e^{Re F (w)} (\cos (\pm π) + i \sin (\pm π)) = - e^{Re F (w)}$ , a real number $< 0$ , contradicting $w \in S$ . Hence $Im F (z) \in (- π, π)$ .

5.1step 3.1step 4.1L4L5L6L14L16

By [L6] there is an integer $k$ with $F (z) = Log z + 2 π i k$ ; taking imaginary parts and using [L4] and [L5], $Im F (z) = θ + 2 π k$ with $- π < θ \leq π$ , and step 4.1 gives $Im F (z) \in (- π, π)$ , so $2 π k \in (- 2 π, 2 π)$ and therefore $k = 0$ by [L16]. Hence $F = Log$ on $S$ .

6.1step 5.1L1L3∎

By step 5.1 the principal logarithm is holomorphic on $S$ , and [L3] applied to $L = Log$ and $h (z) = z$ gives ${Log}^{'} (z) = 1 / z$ there. If $F_{1}$ is any holomorphic function on $S$ with $\exp \circ F_{1} = h$ and $F_{1} (1) = 0$ , then $F_{1} - Log$ is a constant in $2 π i Z$ by [L1], and it vanishes at $1$ , so $F_{1} = Log$ .

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Conventions for chains, cycles and the homological adjective on this page

Remark

Four choices are made in the definitions above, and each is made for a reason that can be stated.

A chain is a finite list, not an element of a group. Complex chains, their traces, and cycles presents a chain as a list of pairs $(m_{k}, γ_{k})$ , with sum given by concatenation and negation by negating the coefficients. Presenting chains as elements of a free abelian group on the set of contours would require saying when two chains are equal, and every such identification would then have to be checked against the integral and the index. The list presentation avoids that obligation entirely: equality of chains is equality of lists, and no result above asserts that two differently presented chains coincide. What the results do assert is equality of the numbers $\int_{Γ} f d z$ and $n (Γ, p)$ , which is all any of them uses.

A cycle is a chain whose boundary function vanishes, and that is weaker than requiring every piece to be closed. In The integral of a continuous derivative over a cycle is zero, summing the endpoint increments of a primitive with the coefficients $m_{k}$ leaves the coefficient of $F (q)$ equal to the boundary value at $q$ . The same boundary cancellation is also used in The index of a cycle about a point off its trace is an integer to make the logarithm increments sum to an integer index. Both arguments need the endpoint counts to cancel and nothing more, so imposing closedness on each $γ_{k}$ would strengthen the hypothesis without strengthening either conclusion. Two contours running between the same pair of distinct points, weighted $+ 1$ and $- 1$ , satisfy the condition and neither is closed.

The adjective is "homologically simply connected", written out every time. Homologically simply connected complex domains defines a condition on indices: every cycle in the domain is null-homologous in it (Null-homologous cycles and homologous cycles in an open set). Nothing above defines or uses a notion of simple connectivity phrased with loops or homotopies, and no statement above asserts a relation between the two. Keeping the qualifier is what makes that scope visible to a reader who arrives with the other notion in mind.

The winding number belongs to the parametrised contour, not to its trace. The winding number of a closed contour about a point off its trace is stated for a map $γ : [a, b] \to C$ , because the integral it is built from depends on that map: a complex contour is a rectifiable path together with its domain and its parametrisation (Rectifiable complex contours, reversal, concatenation, closedness, and orientation), and the trace is only the image set. Two closed contours can share a trace and have different indices at a point, since the parametrisation records how many times and in which direction the trace is traversed; the chain-level index of Integration over a complex chain and the index of a chain inherits the same dependence through its terms.

5 · Examples, counterexamples and false statements

None yet.

The Winding Number and the Global Cauchy Theorem

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Tagged sums approximate a contour integral within oscillation times length

Statement

Facts & Assumptions

Proof

A continuous function holomorphic off a single point is holomorphic

Statement

Facts & Assumptions

Proof

A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic

Statement

Facts & Assumptions

Proof

A holomorphic logarithm is a primitive of the logarithmic derivative

Statement

Facts & Assumptions

Proof

A disc missing pp carries a holomorphic logarithm of z−pz−p

Statement

Facts & Assumptions

Proof

A contour missing a point subdivides into arcs lying in discs that miss it

Statement

Facts & Assumptions

Proof

Continuous logarithms and continuous arguments along a contour

Definition

Remarks

Every contour missing a point admits a continuous logarithm, unique up to a constant in 2πiZ2πiZ

Statement

Facts & Assumptions

Proof

The integral of dz/(z−p)dz/(z−p) along a contour is the increment of a continuous logarithm

Statement

Facts & Assumptions

Proof

The winding number of a closed contour about a point off its trace

Definition

Remarks

The winding number of a closed contour is an integer

Statement

Facts & Assumptions

Proof

The winding number is the increment of a continuous argument divided by 2π2π

Statement

Facts & Assumptions

Proof

The winding number is constant on each connected component of the complement of the trace

Statement

Facts & Assumptions

Proof

The exterior of a closed disc in the plane is path-connected

Statement

Facts & Assumptions

Proof

The complement of a compact plane set has exactly one unbounded connected component

Statement

Facts & Assumptions

Proof

The winding number vanishes on the unbounded component of the complement of the trace

Statement

Facts & Assumptions

Proof

Reversal negates and concatenation adds winding numbers

Statement

Facts & Assumptions

Proof

A circle traversed kk times has winding number kk inside and 00 outside

Statement

Facts & Assumptions

Proof

Complex chains, their traces, and cycles

Definition

Remarks

Integration over a complex chain and the index of a chain

Definition

A disc missing $p$ carries a holomorphic logarithm of $z - p$

Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$

The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm

The winding number is the increment of a continuous argument divided by $2 π$

A circle traversed $k$ times has winding number $k$ inside and $0$ outside