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✓ 11 results · all verified · 7 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 4 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 — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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The unit circle traversed three times has index $3$ at every interior point

Example

Let $γ (t) = \exp (3 i t)$ for $t \in [0, 2 π]$ . Then $γ$ is a closed complex contour with trace the unit circle ${∣ z ∣ = 1}$ , and

$n (γ, z) = 3 for ∣ z ∣ < 1, n (γ, z) = 0 for ∣ z ∣ > 1.$

The function $λ (t) = 3 i t$ is a continuous logarithm of $γ - 0$ along $γ$ , its imaginary part $θ (t) = 3 t$ is a continuous argument running from $0$ to $6 π$ , and $(θ (2 π) - θ (0)) / (2 π) = 3$ recovers the index at the origin.

Facts & Assumptions

Given: The contour $γ (t) = \exp (3 i t)$ on $[0, 2 π]$ .

[L1]

For $a \in C$ , $r > 0$ and $k \in Z$ , the contour $γ_{k} (t) = a + r \exp (i k t)$ on $[0, 2 π]$ is a closed complex contour with $n (γ_{k}, z) = k$ for $∣ z - a ∣ < r$ and $n (γ_{k}, z) = 0$ for $∣ z - a ∣ > r$ ; for $k \neq 0$ its trace is ${z : ∣ z - a ∣ = r}$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L2]

For a closed complex contour $γ$ , a point $p$ off its trace and a continuous argument $θ$ of $γ - p$ along $γ$ , $n (γ, p) = (θ (b) - θ (a)) / (2 π)$ (The winding number is the increment of a continuous argument divided by $2 π$ ).

[L3]

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

[L4]

A continuous logarithm of $γ - p$ along $γ$ is a continuous $λ$ with $\exp (λ (t)) = γ (t) - p$ for every $t$ , and its continuous argument is $Im λ$ (Continuous logarithms and continuous arguments along a contour).

[L5]

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 $e^{i π} + 1 = 0$ ); for $z = a + b i$ with $a, b$ real, $Im z = b$ (Real and imaginary parts, complex conjugation, and modulus).

Verification

technique · direct

1.1

Apply [L1] with $a = 0$ , $r = 1$ and $k = 3$ : the contour is $γ$ , it is a closed complex contour with trace ${∣ z ∣ = 1}$ , and $n (γ, z) = 3$ for $∣ z ∣ < 1$ while $n (γ, z) = 0$ for $∣ z ∣ > 1$ .

givenL1L3

1.2

The map $λ (t) = 3 i t$ is continuous on $[0, 2 π]$ and satisfies $\exp (λ (t)) = \exp (3 i t) = γ (t) - 0$ , so it is a continuous logarithm of $γ - 0$ along $γ$ in the sense of [L4], with continuous argument $θ (t) = Im (3 i t) = 3 t$ by [L5].

givenL4L5

2.1

The argument increment is $θ (2 π) - θ (0) = 6 π - 0 = 6 π$ , so [L2] gives $n (γ, 0) = 6 π / (2 π) = 3$ , the same value step 1.1 assigns at the interior point $0$ .

step 1.1step 1.2L2∎

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A disjoint two-circle cycle has indices $+ 1$ and $- 1$ in its two components

Example

Let $γ_{1} (t) = - 1 + \frac{1}{2} \exp (i t)$ and $γ_{2} (t) = 1 + \frac{1}{2} \exp (- i t)$ for $t \in [0, 2 π]$ , and let $Γ$ be the complex chain $((1, γ_{1}), (1, γ_{2}))$ . Then $Γ$ is a cycle whose trace is the disjoint union of the two circles ${∣ z + 1 ∣ = \frac{1}{2}}$ and ${∣ z - 1 ∣ = \frac{1}{2}}$ , and for $z$ off that trace

$n (Γ, z) = {\begin{cases} + 1, & ∣ z + 1 ∣ < \frac{1}{2}, \\ - 1, & ∣ z - 1 ∣ < \frac{1}{2}, \\ 0, & ∣ z + 1 ∣ > \frac{1}{2} and ∣ z - 1 ∣ > \frac{1}{2} . \end{cases}$

Facts & Assumptions

Given: The contours $γ_{1}, γ_{2}$ above and the chain $Γ = ((1, γ_{1}), (1, γ_{2}))$ .

[L1]

The trace of a sum of chains is the union of their traces; a sum of cycles is a cycle; and for $p$ off the traces involved, $n (Γ_{1} + Γ_{2}, p) = n (Γ_{1}, p) + n (Γ_{2}, p)$ (Chain integration and the index are additive in the chain, and reverse with it).

[L2]

[L3]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ ; a list of closed contours is a cycle; and a single closed contour with coefficient $1$ is a cycle whose trace is the trace of that contour (Complex chains, their traces, and cycles).

[L4]

$n (Γ, p) = (2 π i)^{- 1} \int_{Γ} d z / (z - p)$ , and for a single closed contour with coefficient $1$ this is the winding number of that contour (Integration over a complex chain and the index of a chain).

[L5]

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

Verification

technique · direct

1.1

By [L2] with $a = - 1$ , $r = \frac{1}{2}$ , $k = 1$ the contour $γ_{1}$ is closed with trace ${∣ z + 1 ∣ = \frac{1}{2}}$ , $n (γ_{1}, z) = 1$ for $∣ z + 1 ∣ < \frac{1}{2}$ and $n (γ_{1}, z) = 0$ for $∣ z + 1 ∣ > \frac{1}{2}$ ; by [L2] with $a = 1$ , $r = \frac{1}{2}$ , $k = - 1$ the contour $γ_{2}$ is closed with trace ${∣ z - 1 ∣ = \frac{1}{2}}$ , $n (γ_{2}, z) = - 1$ for $∣ z - 1 ∣ < \frac{1}{2}$ and $n (γ_{2}, z) = 0$ for $∣ z - 1 ∣ > \frac{1}{2}$ .

givenL2

1.2

The two circles are disjoint: if $∣ z + 1 ∣ \leq \frac{1}{2}$ and $∣ z - 1 ∣ \leq \frac{1}{2}$ then $2 = ∣ (z + 1) - (z - 1) ∣ \leq ∣ z + 1 ∣ + ∣ z - 1 ∣ \leq 1$ by [L5], which is false.

2.1

Both contours are closed, so $Γ$ is a cycle by [L3], and by [L1] and [L3] its trace is the union of the two circles, which is disjoint by step 1.2.

step 1.1step 1.2L1L3

3.1

For $z$ off that trace, [L1] and [L4] give $n (Γ, z) = n (γ_{1}, z) + n (γ_{2}, z)$ ; with step 1.1 this is $1 + 0 = 1$ when $∣ z + 1 ∣ < \frac{1}{2}$ , which forces $∣ z - 1 ∣ > \frac{1}{2}$ by step 1.2, and $0 + (- 1) = - 1$ when $∣ z - 1 ∣ < \frac{1}{2}$ , and $0 + 0 = 0$ when both moduli exceed $\frac{1}{2}$ .

step 1.1step 2.1L1L4∎

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The boundary cycle of a round annulus has index $1$ inside the annulus and $0$ on either side

Example

Let $p \in C$ and $0 < r_{1} < r_{2}$ , and let $C_{j} (t) = p + r_{j} \exp (i t)$ on $[0, 2 π]$ for $j = 1, 2$ . Let $Γ$ be the complex chain $((1, C_{2}), (- 1, C_{1}))$ , written $C_{2} - C_{1}$ . Then $Γ$ is a cycle with trace ${∣ z - p ∣ = r_{1}} \cup {∣ z - p ∣ = r_{2}}$ , and

$n (Γ, z) = {\begin{cases} 0, & ∣ z - p ∣ < r_{1}, \\ 1, & r_{1} < ∣ z - p ∣ < r_{2}, \\ 0, & ∣ z - p ∣ > r_{2} . \end{cases}$

Let $s_{1}, s_{2}$ be reals with $0 < s_{1} < r_{1}$ and $r_{2} < s_{2}$ , and put $Ω = {z : s_{1} < ∣ z - p ∣ < s_{2}}$ . Then $Γ$ has trace in the open set $Ω$ and is null-homologous in $Ω$ . The ambient open set is named before the homology because the notion depends on it: the smaller annulus ${r_{1} < ∣ z - p ∣ < r_{2}}$ does not contain the trace of $Γ$ and is therefore not an open set in which $Γ$ is a chain at all.

Facts & Assumptions

Given: A point $p$ , radii $0 < r_{1} < r_{2}$ , the circles $C_{1}, C_{2}$ above, and the chain $Γ = C_{2} - C_{1}$ .

[L1]

The trace of a sum of chains is the union of their traces, the negative of a chain has the same trace, a sum of cycles and the negative of a cycle are cycles, and for $p^{'}$ off the traces involved $n (Γ_{1} + Γ_{2}, p^{'}) = n (Γ_{1}, p^{'}) + n (Γ_{2}, p^{'})$ and $n (- Γ, p^{'}) = - n (Γ, p^{'})$ (Chain integration and the index are additive in the chain, and reverse with it).

[L2]

[L3]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ ; a list of closed contours is a cycle; the negative of a chain negates every coefficient; and a single closed contour with coefficient $1$ is a cycle whose trace is that contour's trace (Complex chains, their traces, and cycles).

[L4]

$n (Γ, p^{'}) = (2 π i)^{- 1} \int_{Γ} d z / (z - p^{'})$ , and for a single closed contour with coefficient $1$ this is that contour's winding number (Integration over a complex chain and the index of a chain).

[L5]

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

Verification

technique · direct

1.1

By [L2] with $k = 1$ each $C_{j}$ is a closed complex contour with trace ${∣ z - p ∣ = r_{j}}$ , with $n (C_{j}, z) = 1$ for $∣ z - p ∣ < r_{j}$ and $n (C_{j}, z) = 0$ for $∣ z - p ∣ > r_{j}$ .

givenL2

2.1

By [L1] and [L3] the chain $Γ = C_{2} - C_{1}$ is a cycle and its trace is ${∣ z - p ∣ = r_{1}} \cup {∣ z - p ∣ = r_{2}}$ , and by [L1] and [L4] its index off that trace is $n (C_{2}, z) - n (C_{1}, z)$ .

step 1.1L1L3L4

3.1

Evaluating step 2.1 with step 1.1: for $∣ z - p ∣ < r_{1} < r_{2}$ the value is $1 - 1 = 0$ ; for $r_{1} < ∣ z - p ∣ < r_{2}$ it is $1 - 0 = 1$ ; for $∣ z - p ∣ > r_{2} > r_{1}$ it is $0 - 0 = 0$ .

step 1.1step 2.1

4.1

With $0 < s_{1} < r_{1}$ and $r_{2} < s_{2}$ the trace of $Γ$ lies in $Ω = {s_{1} < ∣ z - p ∣ < s_{2}}$ , and $C ∖ Ω = {∣ z - p ∣ \leq s_{1}} \cup {∣ z - p ∣ \geq s_{2}}$ ; every point of the first set has $∣ z - p ∣ \leq s_{1} < r_{1}$ and every point of the second has $∣ z - p ∣ \geq s_{2} > r_{2}$ , so step 3.1 gives index $0$ at each, and [L5] makes $Γ$ null-homologous in $Ω$ .

step 2.1step 3.1L5∎

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The winding numbers of a keyhole contour about the origin and about an excluded point

Example

Let $0 < ε < R$ and put

$C_{R} (t) = R \exp (i t), C_{ε}^{-} (t) = ε \exp (- i t) (t \in [0, 2 π]),$ $σ_{1} (t) = ε + t (R - ε), σ_{2} (t) = R + t (ε - R) (t \in [0, 1]) .$

The keyhole is the complex chain $Γ = ((1, C_{R}), (1, C_{ε}^{-}), (1, σ_{1}), (1, σ_{2}))$ . Then $Γ$ is a cycle, its trace is

$Γ^{*} = {∣ z ∣ = R} \cup {∣ z ∣ = ε} \cup {x \in R : ε \leq x \leq R},$

and at every point $z$ off that trace

$n (Γ, z) = {\begin{cases} 0, & ∣ z ∣ < ε, \\ 1, & ε < ∣ z ∣ < R, \\ 0, & ∣ z ∣ > R . \end{cases}$

The two radial segments have the same trace and both carry coefficient $1$ , so the closed segment from $ε$ to $R$ on the real axis belongs to $Γ^{*}$ and no index is asserted at any of its points.

Facts & Assumptions

Given: Reals $0 < ε < R$ and the four contours above forming the chain $Γ$ .

[L1]

The trace of a sum of chains is the union of their traces, a sum of cycles is a cycle, and for $q$ off the traces involved $n (Γ_{1} + Γ_{2}, q) = n (Γ_{1}, q) + n (Γ_{2}, q)$ and $n (Γ^{-}, q) = - n (Γ, q)$ , where $Γ^{-}$ reverses every contour (Chain integration and the index are additive in the chain, and reverse with it).

[L2]

[L3]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ ; 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; and a list of closed contours is a cycle (Complex chains, their traces, and cycles).

[L4]

$n (Γ, q) = (2 π i)^{- 1} \int_{Γ} d z / (z - q)$ for $q$ off the trace, and for a single closed contour with coefficient $1$ this is that contour's winding number (Integration over a complex chain and the index of a chain).

[L5]

The reversal of $γ : [a, b] \to C$ is $γ^{-} (t) = γ (a + b - t)$ , and it is again a complex contour with the same trace (Reversal negates and concatenation adds winding numbers, Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

[L6]

For $p \in C$ and $0 < r_{1} < r_{2}$ , the chain $C_{2} - C_{1}$ built from the positively oriented circles of radii $r_{1}, r_{2}$ about $p$ has index $0$ for $∣ z - p ∣ < r_{1}$ , $1$ for $r_{1} < ∣ z - p ∣ < r_{2}$ and $0$ for $∣ z - p ∣ > r_{2}$ (The boundary cycle of a round annulus has index $1$ inside the annulus and $0$ on either side).

[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).

Verification

technique · direct

1.1

The segments $σ_{1}, σ_{2}$ are affine, hence rectifiable by [L7], with $σ_{1} (0) = ε$ , $σ_{1} (1) = R$ , $σ_{2} (0) = R$ , $σ_{2} (1) = ε$ , and both have trace the closed real segment from $ε$ to $R$ ; the two circles are closed complex contours by [L2], with traces ${∣ z ∣ = R}$ and ${∣ z ∣ = ε}$ .

givenL2L7

1.2

$σ_{2}$ is the reversal of $σ_{1}$ : $σ_{1}^{-} (t) = σ_{1} (1 - t) = ε + (1 - t) (R - ε) = R + t (ε - R) = σ_{2} (t)$ , so by [L1] and [L5] the one-term chains $((1, σ_{1}))$ and $((1, σ_{2}))$ have indices summing to $0$ at every point off the segment.

givenL1L5

2.1

$Γ$ is a cycle: by [L3] the two closed circles contribute nothing to $\partial Γ$ , while $σ_{1}$ contributes $+ 1$ at $R$ and $- 1$ at $ε$ and $σ_{2}$ contributes $+ 1$ at $ε$ and $- 1$ at $R$ , so every value of $\partial Γ$ is $0$ . Its trace is the union named in the statement, by step 1.1 and [L1].

step 1.1L1L3

3.1

For $z \notin Γ^{*}$ , [L1] and [L4] split the index into the four one-term contributions, of which the two segment terms cancel by step 1.2; so $n (Γ, z) = n (C_{R}, z) + n (C_{ε}^{-}, z)$ , which by [L2] is $1 + (- 1) = 0$ for $∣ z ∣ < ε$ , $1 + 0 = 1$ for $ε < ∣ z ∣ < R$ , and $0 + 0 = 0$ for $∣ z ∣ > R$ . The same three values are what [L6] gives for the annulus cycle built from the positively oriented circles of radii $ε$ and $R$ about $0$ ; that chain is a different list from $Γ$ , and what is asserted here is only that the two index functions agree off the traces.

step 1.2step 2.1L1L2L4L6∎

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A continuous argument computed along a spiralling contour

Example

Let $γ (t) = (1 + t) \exp (2 π i t)$ for $t \in [0, 1]$ and let $p = 0$ . Then $γ$ is a complex contour with $0 \notin γ^{*}$ , and

$λ (t) = \log (1 + t) + 2 π i t$

is a continuous logarithm of $γ - 0$ along $γ$ , with continuous argument $θ (t) = 2 π t$ increasing by $2 π$ . Consequently

$\int_{γ} \frac{d z}{z} = λ (1) - λ (0) = \log 2 + 2 π i .$

The contour is not closed: $γ (0) = 1$ and $γ (1) = 2$ . So no winding number is defined for it, and the increment $\log 2 + 2 π i$ is not an element of $2 π i Z$ . This is exactly the gap between the logarithm-increment identity, which holds for every contour missing $p$ , and the integrality statement, which needs closedness.

Facts & Assumptions

Given: The contour $γ (t) = (1 + t) \exp (2 π i t)$ on $[0, 1]$ and the point $p = 0$ .

[L1]

[L2]

For a complex contour $γ$ and $p \notin γ^{*}$ there is a continuous logarithm of $γ - p$ along $γ$ , and any two differ by a constant 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 $γ : [a, b] \to C$ , 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).

[L4]

For $x > 0$ , $\log x$ is the unique real $y$ with $\exp y = x$ (The natural logarithm as the inverse of the exponential function); the natural logarithm is continuous and strictly increasing on $(0, \infty)$ , and $\log 1 = 0$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L5]

A complex contour is a rectifiable path, and it is closed when its two endpoint values agree (Rectifiable complex contours, reversal, concatenation, closedness, and orientation); 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).

[L6]

$\exp (z + w) = \exp z \exp w$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential), and for real $x, y$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ with $∣ \exp (x + i y) ∣ = e^{x}$ ; in particular $\exp (2 π i) = 1$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Quarter-turn values and shifts by pi/2 and pi).

[L7]

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

[L8]

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

Verification

technique · direct

1.1

Writing $γ (t) = (1 + t) \cos (2 π t) + i (1 + t) \sin (2 π t)$ by [L6], the path is differentiable in $t$ with a continuous derivative by [L8], hence rectifiable by [L5]; and $∣ γ (t) ∣ = (1 + t) ∣ \exp (2 π i t) ∣ = 1 + t \geq 1$ by [L6], so $0$ does not lie on the trace.

givenL5L6L8

1.2

The map $λ (t) = \log (1 + t) + 2 π i t$ is continuous on $[0, 1]$ , and by [L4] and [L6], $\exp (λ (t)) = e^{\log (1 + t)} \exp (2 π i t) = (1 + t) \exp (2 π i t) = γ (t) - 0$ ; so $λ$ is a continuous logarithm of $γ - 0$ along $γ$ in the sense of [L1], with continuous argument $θ (t) = Im λ (t) = 2 π t$ by [L7].

givenL1L4L6L7

2.1

By [L3] and step 1.2, $\int_{γ} d z / z = λ (1) - λ (0) = (\log 2 + 2 π i) - (\log 1 + 0) = \log 2 + 2 π i$ using [L4]; and $θ (1) - θ (0) = 2 π$ . By [L2] the value does not depend on which continuous logarithm is taken.

step 1.1step 1.2L2L3L4

3.1

The endpoint values are $γ (0) = 1$ and $γ (1) = 2 \exp (2 π i) = 2$ by [L6], so $γ$ is not closed by [L5] and no winding number is defined for it; consistently, $\log 2 + 2 π i$ is not an element of $2 π i Z$ because $\log 2 \neq 0$ by [L4].

step 2.1L4L5L6∎

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Dixon's gluing traced on the boundary cycle of an annulus

Example

Let $Ω = {z : \frac{1}{2} < ∣ z ∣ < 3}$ , let $C_{1} (t) = \exp (i t)$ and $C_{2} (t) = 2 \exp (i t)$ on $[0, 2 π]$ , let $Γ = C_{2} - C_{1}$ and let $f (z) = 1 / z$ , holomorphic on $Ω$ . Then $Γ$ is a cycle with trace ${∣ z ∣ = 1} \cup {∣ z ∣ = 2} \subseteq Ω$ , null-homologous in $Ω$ , and

$Ω_{0} = {z \notin Γ^{*} : n (Γ, z) = 0} = {∣ z ∣ < 1} \cup {∣ z ∣ > 2}, Ω \cup Ω_{0} = C .$

Dixon's glued function is identically zero here: the transform

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

vanishes at every $z \in Ω_{0}$ , by direct computation and not only by Liouville's theorem. At $z = \frac{3}{2}$ , which lies in $Ω$ off the trace, both sides of the global Cauchy formula equal $\frac{2}{3}$ .

Facts & Assumptions

Given: The sets and contours above, with $f (z) = 1 / z$ .

[L1]

If $Ω$ is open, $f$ is holomorphic on $Ω$ , and $Γ$ is a null-homologous cycle with trace in $Ω$ , then, with $g$ the filled difference quotient of $f$ , the function 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 $Ω_{0}$ is a well-defined entire function, bounded and tending to $0$ at infinity (Dixon's glued function is entire and vanishes at infinity). The filled difference quotient is $(f (ζ) - f (z)) / (ζ - z)$ off the diagonal and $f^{'} (z)$ on it (The filled difference quotient of a holomorphic function is jointly continuous).

[L2]

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

[L3]

For $p \in C$ and $0 < r_{1} < r_{2}$ the chain built from the positively oriented circles of radii $r_{2}$ and $r_{1}$ about $p$ with coefficients $+ 1$ and $- 1$ is a cycle with trace the two circles, index $0$ for $∣ z - p ∣ < r_{1}$ , $1$ for $r_{1} < ∣ z - p ∣ < r_{2}$ and $0$ for $∣ z - p ∣ > r_{2}$ ; it is null-homologous in ${s_{1} < ∣ z - p ∣ < s_{2}}$ whenever $0 < s_{1} < r_{1}$ and $r_{2} < s_{2}$ (The boundary cycle of a round annulus has index $1$ inside the annulus and $0$ on either side).

[L4]

A cycle with trace in an open $Ω$ is null-homologous in $Ω$ when its index vanishes at every point outside $Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L5]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ , 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 (Complex chains, their traces, and cycles).

[L6]

For $a \in C$ , $r > 0$ and every integer $m$ , the positively oriented circle $γ (t) = a + r \exp (i t)$ on $[0, 2 π]$ satisfies $\int_{γ} (z - a)^{m} d z = 2 π i$ when $m = - 1$ and $0$ otherwise (On a positively oriented circle about a, the integral of (z-a)^m is zero for every integer m except -1, and is 2 pi i for m=-1).

[L7]

Chain integration and the index are additive in the chain and reverse with it (Chain integration and the index are additive in the chain, and reverse with it); complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand).

[L8]

For $a \in C$ , $r > 0$ and $k \in Z$ , the contour $a + r \exp (i k t)$ on $[0, 2 π]$ has index $k$ for $∣ z - a ∣ < r$ and $0$ for $∣ z - a ∣ > r$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L9]

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

[L10]

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

Verification

technique · direct

1.1

By [L3] with $p = 0$ , $r_{1} = 1$ , $r_{2} = 2$ , $s_{1} = \frac{1}{2}$ and $s_{2} = 3$ , the chain $Γ$ is a cycle with trace ${∣ z ∣ = 1} \cup {∣ z ∣ = 2}$ contained in $Ω$ , its index is $0$ for $∣ z ∣ < 1$ , $1$ for $1 < ∣ z ∣ < 2$ and $0$ for $∣ z ∣ > 2$ , and it is null-homologous in $Ω$ . Hence $Ω_{0} = {∣ z ∣ < 1} \cup {∣ z ∣ > 2}$ , and $Ω \cup Ω_{0} = C$ because a point with $∣ z ∣ \leq \frac{1}{2}$ or $∣ z ∣ \geq 3$ lies in $Ω_{0}$ .

givenL3L4L5

1.2

The function $f (z) = 1 / z$ is holomorphic on $Ω$ by [L10], since $0 \notin Ω$ .

givenL10

2.1

For $z \in Ω_{0}$ with $z \neq 0$ and $ζ$ on either circle, the identity $\frac{1}{ζ (ζ - z)} = \frac{1}{z} (\frac{1}{ζ - z} - \frac{1}{ζ})$ holds, and [L5], [L6] and [L8] give $\int_{C_{j}} \frac{d ζ}{ζ - z} = 2 π i n (C_{j}, z)$ and $\int_{C_{j}} \frac{d ζ}{ζ} = 2 π i$ . For $∣ z ∣ < 1$ both indices are $1$ , so each circle integral is $\frac{1}{z} (2 π i - 2 π i) = 0$ ; for $∣ z ∣ > 2$ both indices are $0$ , so each is $\frac{1}{z} (0 - 2 π i) = - 2 π i / z$ . In both cases [L7] gives $h_{1} (z) = 0$ as the difference of the two equal circle contributions.

step 1.1step 1.2L5L6L7L8algebra

2.2

At $z = 0$ the integrand is $ζ^{- 2}$ , and [L6] with $m = - 2$ gives $\int_{C_{j}} ζ^{- 2} d ζ = 0$ for both circles, so $h_{1} (0) = 0$ as well.

step 1.1L5L6

3.1

Steps 2.1 and 2.2 give $h_{1} \equiv 0$ on $Ω_{0}$ ; by [L1] the glued function agrees with $h_{1}$ there and is entire and bounded, so [L9] makes it the constant $0$ , and the value on $Ω$ is therefore $0$ too.

step 2.1step 2.2L1L9

4.1

Take $z = \frac{3}{2}$ , so $1 < ∣ z ∣ < 2$ and $z \in Ω ∖ Γ^{*}$ . The left side of [L2] is $n (Γ, z) f (z) = 1 \cdot \frac{2}{3} = \frac{2}{3}$ by step 1.1. For the right side, step 2.1's partial-fraction identity with $n (C_{2}, z) = 1$ and $n (C_{1}, z) = 0$ from [L8] gives $\int_{C_{2}} \frac{d ζ}{ζ (ζ - z)} = \frac{1}{z} (2 π i - 2 π i) = 0$ and $\int_{C_{1}} \frac{d ζ}{ζ (ζ - z)} = \frac{1}{z} (0 - 2 π i) = - 2 π i / z$ , so by [L7] the right side is $\frac{1}{2 π i} (0 + 2 π i / z) = \frac{1}{z} = \frac{2}{3}$ . The two sides agree.

step 1.1step 2.1step 3.1L2L6L7L8algebra∎

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Every cycle in a round annulus has one period, that of the central circle

Example

Let $0 < r_{1} < r_{2}$ , let $A = {z : r_{1} < ∣ z ∣ < r_{2}}$ , and let $C (t) = ρ \exp (i t)$ on $[0, 2 π]$ for a fixed $ρ$ with $r_{1} < ρ < r_{2}$ . Let $Γ$ be any complex chain which is a cycle with trace in $A$ , and put $k = n (Γ, 0)$ , an integer. Then $Γ$ and the chain $k C$ , consisting of $C$ with coefficient $k$ , are homologous in $A$ , and consequently

$\int_{Γ} f (z) d z = k \int_{C} f (z) d z$

for every holomorphic $f$ on $A$ . For $f (z) = 1 / z$ the right-hand factor is $\int_{C} d z / z = 2 π i$ , so $\int_{Γ} d z / z = 2 π i k$ .

Facts & Assumptions

Given: Radii $0 < r_{1} < ρ < r_{2}$ , the annulus $A$ , the circle $C$ , and a cycle $Γ$ with trace in $A$ .

[L1]

If $f$ is holomorphic on an open $Ω$ and two cycles with traces in $Ω$ are homologous in $Ω$ , their integrals of $f$ agree (Holomorphic integrals agree on homologous cycles).

[L2]

Two cycles with traces in $Ω$ are homologous in $Ω$ exactly when their indices agree at every point of $C ∖ Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L3]

For $a \in C$ , $r > 0$ and $k^{'} \in Z$ , the contour $a + r \exp (i k^{'} t)$ on $[0, 2 π]$ has index $k^{'}$ for $∣ z - a ∣ < r$ and $0$ for $∣ z - a ∣ > r$ , with trace ${∣ z - a ∣ = r}$ when $k^{'} \neq 0$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L5]

For a cycle $Γ$ the trace is compact, the index is constant on every connected component of $C ∖ Γ^{*}$ , each such component is open, and there is $R > 0$ with $n (Γ, p) = 0$ whenever $∣ p ∣ > R$ (The index of a cycle is locally constant off its trace and vanishes far from it).

[L6]

For $a \in C$ , $r > 0$ and every integer $m$ , the positively oriented circle $a + r \exp (i t)$ on $[0, 2 π]$ satisfies $\int_{γ} (z - a)^{m} d z = 2 π i$ when $m = - 1$ and $0$ otherwise (On a positively oriented circle about a, the integral of (z-a)^m is zero for every integer m except -1, and is 2 pi i for m=-1).

[L7]

$\int_{Γ} f d z = \sum_{k^{'} < r, m_{k^{'}} \neq 0} m_{k^{'}} \int_{γ_{k^{'}}} f d z$ and $n (Γ, p) = (2 π i)^{- 1} \int_{Γ} d z / (z - p)$ (Integration over a complex chain and the index of a chain), a chain being a finite list of integer-weighted contours, and a one-term chain carried by a closed contour being a cycle (Complex chains, their traces, and cycles).

[L8]

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

[L9]

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

[L10]

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).

[L11]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ 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); 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).

[L12]

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).

[L13]

Constants and the identity are holomorphic, and nonvanishing quotients of holomorphic functions are holomorphic (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Verification

technique · direct

1.1

The trace of $Γ$ lies in $A$ , so it misses the closed disc $D = {∣ z ∣ \leq r_{1}}$ and the closed exterior $E = {∣ z ∣ \geq r_{2}}$ , whose union is $C ∖ A$ . The number $k = n (Γ, 0)$ is defined because $0 \notin A$ , and it is an integer by [L12].

givenL7L12

1.2

By [L3] the contour $C$ is closed, with index $1$ on $D$ and $0$ on $E$ . Since $k C$ is the one-term chain carrying $C$ with coefficient $k$ , [L7] gives $n (k C, p) = k n (C, p)$ at every $p$ off ${∣ z ∣ = ρ}$ , so it is $k$ on $D$ , since $∣ p ∣ \leq r_{1} < ρ$ there, and $0$ on $E$ , since $∣ p ∣ \geq r_{2} > ρ$ there; its trace is contained in ${∣ z ∣ = ρ} \subseteq A$ , and it is a cycle by [L7].

givenL3L7

2.1

$D$ is convex by [L10] and [L11], hence connected, and $E$ is connected by [L8]; both are subsets of $C ∖ Γ^{*}$ by step 1.1.

step 1.1L8L10L11

3.1

By [L9] the connected set $D$ lies inside a single component of $C ∖ Γ^{*}$ , on which the index is constant by [L5]; since $0 \in D$ , this gives $n (Γ, p) = k$ for every $p \in D$ .

step 2.1L5L9

3.2

By [L9] the connected set $E$ lies inside a single component of $C ∖ Γ^{*}$ ; by [L5] there is $R > 0$ with $n (Γ, q) = 0$ whenever $∣ q ∣ > R$ , and $E$ contains the point $\max (r_{2}, R) + 1$ of modulus greater than $R$ , so the constant value of the index on that component is $0$ : thus $n (Γ, p) = 0$ for every $p \in E$ .

step 2.1L5L9L11

4.1

Steps 3.1, 3.2 and 1.2 make the indices of $Γ$ and $k C$ agree at every point of $C ∖ A = D \cup E$ , so [L2] makes them homologous in the open set $A$ , and [L1] gives $\int_{Γ} f d z = \int_{k C} f d z = k \int_{C} f d z$ for every holomorphic $f$ on $A$ , the last equality by [L7].

step 3.1step 3.2step 1.2L1L2L7

5.1

The identity map $z \mapsto z$ is holomorphic and nonvanishing on $A$ , because $0 \notin A$ ; hence [L13] makes $f (z) = 1 / z$ holomorphic on $A$ . Then [L6] with $a = 0$ and $m = - 1$ gives $\int_{C} d z / z = 2 π i$ , so step 4.1 yields $\int_{Γ} d z / z = 2 π i k$ .

step 4.1L6L13∎

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The winding number is the circulation of the planar vortex field divided by $2 π$

Remark

The hypothesis of this remark is piecewise $C^{1}$ , not merely rectifiable. The real line integrals it quotes are those of Scalar line integrals with respect to arc length and vector-field line integrals, which are defined through a derivative of the path; a general complex contour is only rectifiable and has no derivative, so the identification below is asserted only for a piecewise- $C^{1}$ closed contour $γ : [a, b] \to C ∖ {0}$ . Read the plane as $R^{2}$ through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves and write $γ = (x, y)$ for the corresponding planar path.

The split. With $Re$ , $Im$ and the modulus as in Real and imaginary parts, complex conjugation, and modulus, the identity $1 / z = \overline{z} / ∣ z ∣^{2}$ turns the integrand into

$\frac{d z}{z} = \frac{(x - i y) (d x + i d y)}{x^{2} + y^{2}} = \frac{x d x + y d y}{x^{2} + y^{2}} + i \frac{x d y - y d x}{x^{2} + y^{2}},$

and For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals identifies the two real parts of $\int_{γ} d z / z$ with the vector line integrals of

$P (x, y) = (\frac{x}{x^{2} + y^{2}}, \frac{y}{x^{2} + y^{2}}), V (x, y) = (\frac{- y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}})$

along $γ$ .

The first field contributes nothing. $P$ is the gradient of $ϕ (x, y) = \frac{1}{2} \log (x^{2} + y^{2})$ on the punctured plane, so The gradient theorem: the line integral of a gradient is the endpoint increment evaluates its line integral as $ϕ (γ (b)) - ϕ (γ (a))$ , which is $0$ for a closed path; the same conclusion is what Conservative fields are path-independent and have zero integral around every closed path records for a conservative field.

The second field is the vortex field. $V$ is the $C^{1}$ field on $R^{2} ∖ {(0, 0)}$ whose partial derivatives satisfy the closedness condition of Exact and closed C1 vector fields while admitting no global potential there. Since $\int_{γ} \frac{d z}{z} = 2 π i n (γ, 0)$ by The winding number of a closed contour about a point off its trace, the circulation of $V$ around $γ$ is $2 π n (γ, 0)$ .

What the two statements share. A nonzero winding number and the failure of $V$ to be exact on the punctured plane are the same fact recorded in two vocabularies: if $V$ had a $C^{2}$ potential on the punctured plane, then The gradient theorem: the line integral of a gradient is the endpoint increment would force its circulation, and hence $n (γ, 0)$ , to vanish around every closed piecewise- $C^{1}$ path there; A circle traversed $k$ times has winding number $k$ inside and $0$ outside exhibits circles with index $k$ about the origin for every integer $k$ . That exactness needs more than closedness on a domain of this shape is the point recorded in Closedness is local, exactness is global, and a domain hypothesis cannot be omitted, and Every exact C1 vector field is closed is the implication that does hold on every open set.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A connected plane domain that is not homologically simply connected

Statement refuted

Every complex domain is homologically simply connected.

Facts & Assumptions

Given: The annulus $A = {z \in C : \frac{1}{2} < ∣ z ∣ < 2}$ and the contour $C (t) = \exp (i t)$ on $[0, 2 π]$ , taken as the chain with the single term $(1, C)$ .

[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).

[L2]

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

[L3]

For $a \in C$ , $r > 0$ and $k \in Z$ , the contour $a + r \exp (i k t)$ on $[0, 2 π]$ is a closed complex contour with index $k$ for $∣ z - a ∣ < r$ and $0$ for $∣ z - a ∣ > r$ , and with trace ${∣ z - a ∣ = r}$ when $k \neq 0$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L4]

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

[L5]

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

[L6]

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

[L7]

A single closed contour with coefficient $1$ is a cycle whose trace is that contour's trace (Complex chains, their traces, and cycles), and its index is the winding number of that contour (Integration over a complex chain and the index of a chain).

[L8]

For $n \geq 2$ the unit sphere $S^{n - 1} \subseteq R^{n}$ is path-connected (For $n \geq 2$ , the sphere $S^{n - 1}$ is path-connected and connected); the map $x \mapsto x / ∥ x ∥_{2}$ into $S^{n - 1}$ is continuous (Radial normalisation $x \mapsto x / ∥ x ∥_{2}$ is continuous on $R^{n} ∖ {0}$ ); and $S^{n - 1} = {x : ∥ x ∥_{2} = 1}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[L9]

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

[L10]

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

[L11]

A set is open exactly when each of its points admits a ball inside it, and $B (x, r) = {y : d (x, y) < r}$ (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).

Counterexample

technique · constructive

1.1

Take $A = {\frac{1}{2} < ∣ z ∣ < 2}$ and $C (t) = \exp (i t)$ on $[0, 2 π]$ .

givenconstruct

1.2

$A$ is open: for $z \in A$ the ball of radius $\min (∣ z ∣ - \frac{1}{2}, 2 - ∣ z ∣)$ about $z$ lies in $A$ by [L10] and [L11]; and $A$ is nonempty, since $1 \in A$ .

givenL10L11

2.1

$A$ is path-connected, hence connected by [L9]: given $z, w \in A$ , the ambient exterior ${∣ u ∣ > \frac{1}{2}}$ is path-connected by [L6], and more concretely the radial paths $s \mapsto ((1 - s) ∣ z ∣ + s) z / ∣ z ∣$ and $s \mapsto ((1 - s) ∣ w ∣ + s) w / ∣ w ∣$ are continuous by [L8] and [L9], keep the modulus between $1$ and the starting modulus, hence inside $(\frac{1}{2}, 2)$ , and end on the unit circle; the unit circle is path-connected by [L8], and its points have modulus $1$ ; concatenating the three pieces on closed subintervals of $[0, 1]$ gives a continuous path in $A$ from $z$ to $w$ by [L9].

step 1.1L6L8L9L10

2.2

By [L3] with $a = 0$ , $r = 1$ , $k = 1$ the contour $C$ is closed with trace ${∣ z ∣ = 1} \subseteq A$ and $n (C, 0) = 1$ , a value [L4] confirms directly; by [L7] the chain with the single term $(1, C)$ is a cycle with the same trace and the same index.

step 1.1L3L4L7

3.1

Steps 1.2 and 2.1 make $A$ a complex domain in the sense of [L5]. The point $0$ lies in $C ∖ A$ , and step 2.2 gives $n (C, 0) = 1 \neq 0$ , so by [L2] the cycle $C$ is not null-homologous in $A$ ; by [L1] the domain $A$ is therefore not homologically simply connected, which refutes the claim.

step 1.2step 2.1step 2.2L1L2L5discharge-construct∎

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A nonvanishing holomorphic function on a domain with no holomorphic logarithm

Statement refuted

Every holomorphic nowhere-zero function on a complex domain has a holomorphic logarithm on that domain.

Facts & Assumptions

Given: The punctured plane $U = C ∖ {0}$ , the identity function $f (z) = z$ on it, and the contour $C (t) = \exp (i t)$ on $[0, 2 π]$ .

[L1]

On a homologically simply connected complex domain, a holomorphic nowhere-zero function admits a holomorphic $L$ with $\exp \circ L$ equal to it (A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm), a domain being homologically simply connected when every cycle in it is null-homologous in it (Homologically simply connected complex domains).

[L2]

There is no continuous $L : C ∖ {0} \to C$ with $\exp (L (z)) = z$ for every $z \neq 0$ (There is no continuous logarithm on all of $C ∖ {0}$ ).

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

[L5]

If $F$ is holomorphic on an open set, $F^{'}$ is continuous there, and $γ$ is a closed rectifiable contour in that set, then $\int_{γ} F^{'} (z) d z = 0$ (The integral of a continuous complex derivative over every closed rectifiable contour is zero).

[L6]

A cycle with trace in an open $Ω$ is null-homologous in $Ω$ when its index vanishes at every point of $C ∖ Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L7]

The annulus ${\frac{1}{2} < ∣ z ∣ < 2}$ is a complex domain that is not homologically simply connected, the unit circle in it having index $1$ about the origin (A connected plane domain that is not homologically simply connected).

[L8]

For $a \in C$ , $r > 0$ and $k \in Z$ , the contour $a + r \exp (i k t)$ on $[0, 2 π]$ is a closed complex contour with index $k$ for $∣ z - a ∣ < r$ and $0$ for $∣ z - a ∣ > r$ , with trace ${∣ z - a ∣ = r}$ when $k \neq 0$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L9]

A complex domain is a nonempty, connected, open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ); for $c \in C$ and $R \geq 0$ the set ${z : ∣ z - c ∣ > R}$ is path-connected and connected, and at $R = 0$ this is the punctured plane (The exterior of a closed disc in the plane is path-connected).

[L10]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there), and linear combinations, products and nonvanishing quotients of complex differentiable functions are complex differentiable, the identity having derivative $1$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L11]

A single closed contour with coefficient $1$ is a cycle whose trace is that contour's trace (Complex chains, their traces, and cycles) and whose index is that contour's winding number (Integration over a complex chain and the index of a chain).

[L12]

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).

Counterexample

technique · constructive

1.1

Take $U = C ∖ {0}$ and $f (z) = z$ on it.

givenconstruct

2.1

$U$ is a complex domain: it is nonempty, open by [L12], and connected by [L9] with $c = 0$ and $R = 0$ . The function $f$ is holomorphic on $U$ by [L10] and nowhere zero there, since $0 \notin U$ .

step 1.1L9L10L12

3.1

Suppose $L$ were a holomorphic function on $U$ with $\exp (L (z)) = z$ for every $z \in U$ . Then $L$ is continuous on $U$ by [L10], contradicting [L2]; so no such $L$ exists and the claim is refuted.

step 2.1L2L10

4.1

A second refutation, independent of [L2]. With $L$ as in step 3.1, [L3] gives $L^{'} (z) = 1 / z$ , which is continuous on $U$ by [L10], so [L5] applied to the closed rectifiable contour $C$ in $U$ gives $\int_{C} d z / z = \int_{C} L^{'} (z) d z = 0$ ; but [L4] gives $\int_{C} d z / z = 2 π i \neq 0$ .

step 2.1L3L4L5L10

5.1

The hypothesis of [L1] that fails is homological simple connectivity: by [L8] and [L11] the unit circle $C$ is a cycle with trace in $U$ and $n (C, 0) = 1$ , while $0 \in C ∖ U$ , so $C$ is not null-homologous in $U$ by [L6]. The same failure on the smaller annulus is recorded in [L7].

step 3.1step 4.1L1L6L7L8L11discharge-construct∎

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

The winding number depends only on the trace of the closed contour

Statement

False claim. If two closed complex contours have the same trace, then they have the same winding number about every point off that trace.

Facts & Assumptions

Given: The contours $γ_{1} (t) = \exp (i t)$ and $γ_{2} (t) = \exp (2 i t)$ on $[0, 2 π]$ .

[L1]

[L2]

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

[L3]

A complex contour is a rectifiable path together with its parameter interval and its parametrisation; its trace is only the image set (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

Refutation

technique · direct

1.1

By [L1] with $a = 0$ , $r = 1$ and $k = 1$ , the contour $γ_{1}$ is closed with trace ${∣ z ∣ = 1}$ and $n (γ_{1}, 0) = 1$ .

givenL1

1.2

By [L1] with $a = 0$ , $r = 1$ and $k = 2$ , the contour $γ_{2}$ is closed with trace ${∣ z ∣ = 1}$ and $n (γ_{2}, 0) = 2$ .

givenL1

2.1

The two contours have the same trace, and $0$ lies off it, yet $1 \neq 2$ ; so the claim is false.

step 1.1step 1.2

3.1

Nothing here is anomalous: by [L2] the index is computed from an integral over the parametrised contour, and by [L3] the trace forgets the parametrisation, which is what records how many times the circle is traversed. This is why The winding number of a closed contour about a point off its trace attaches the index to the map and not to the image set.

step 2.1L2L3∎

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

Every cycle in a connected plane domain is null-homologous in that domain

Statement

False claim. Every complex chain which is a cycle with trace in a complex domain is null-homologous in that domain.

Facts & Assumptions

Given: The annulus $A = {z : \frac{1}{2} < ∣ z ∣ < 2}$ and the contour $C (t) = \exp (i t)$ on $[0, 2 π]$ , taken as the chain with the single term $(1, C)$ .

[L1]

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

[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).

[L3]

The annulus ${\frac{1}{2} < ∣ z ∣ < 2}$ is a complex domain, and the unit circle $C$ in it is a cycle with $n (C, 0) = 1$ , so $C$ is not null-homologous in that annulus (A connected plane domain that is not homologically simply connected).

[L4]

For $a \in C$ , $r > 0$ and $k \in Z$ , the contour $a + r \exp (i k t)$ on $[0, 2 π]$ is a closed complex contour with index $k$ for $∣ z - a ∣ < r$ and $0$ for $∣ z - a ∣ > r$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

[L5]

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).

[L6]

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

[L7]

[L8]

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

Refutation

technique · direct

1.1

By [L3] the annulus $A$ is a complex domain and the chain $C$ is a cycle with trace ${∣ z ∣ = 1} \subseteq A$ ; by [L4] and [L7] its index at $0$ is $1$ .

givenL3L4L7

2.1

The point $0$ lies in $C ∖ A$ , and step 1.1 gives $n (C, 0) = 1 \neq 0$ , so [L1] denies that $C$ is null-homologous in $A$ ; the claim is therefore false, and by [L2] it is exactly the claim that every complex domain is homologically simply connected.

step 1.1L1L2

3.1

The hypothesis is not removable from Cauchy's theorem either: $f (z) = 1 / z$ is holomorphic on $A$ by [L8], since $0 \notin A$ , and $\int_{C} d z / z = 2 π i \neq 0$ by [L6], so the conclusion of [L5] fails for this cycle in this domain.

step 2.1L5L6L8∎

The Winding Number and the Global Cauchy Theorem — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The unit circle traversed three times has index 33 at every interior point

Example

Facts & Assumptions

Verification

A disjoint two-circle cycle has indices +1+1 and −1−1 in its two components

Example

Facts & Assumptions

Verification

The boundary cycle of a round annulus has index 11 inside the annulus and 00 on either side

Example

Facts & Assumptions

Verification

The winding numbers of a keyhole contour about the origin and about an excluded point

Example

Facts & Assumptions

Verification

A continuous argument computed along a spiralling contour

Example

Facts & Assumptions

Verification

Dixon's gluing traced on the boundary cycle of an annulus

Example

Facts & Assumptions

Verification

Every cycle in a round annulus has one period, that of the central circle

Example

Facts & Assumptions

Verification

The winding number is the circulation of the planar vortex field divided by 2π2π

Remark

A connected plane domain that is not homologically simply connected

Statement refuted

Facts & Assumptions

Counterexample

A nonvanishing holomorphic function on a domain with no holomorphic logarithm

Statement refuted

Facts & Assumptions

Counterexample

The winding number depends only on the trace of the closed contour

Statement

Facts & Assumptions

Refutation

Every cycle in a connected plane domain is null-homologous in that domain

Statement

Facts & Assumptions

Refutation

Sources

The unit circle traversed three times has index $3$ at every interior point

A disjoint two-circle cycle has indices $+ 1$ and $- 1$ in its two components

The boundary cycle of a round annulus has index $1$ inside the annulus and $0$ on either side

The winding number is the circulation of the planar vortex field divided by $2 π$