TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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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]

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).

[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 $γ$ .

Depends on

Used by

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §2.1, Lemma 1 (standard reference, not scraped)
J. Lebl, Complex Analysis, Ch. 4 §4.1 (standard reference, not scraped)
M. Weber, Complex Analysis (Indiana University), Ch. 4 §4.1 (standard reference, not scraped)