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

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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]

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).

[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]

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]

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 $R$ ").

[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]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ 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]

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

1.1givenL9L10

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 $C ∖ γ^{*}$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources