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CorollaryStatement: 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 the increment of a continuous argument divided by 2π

Statement

Let γ:[a,b]→C be a closed complex contour, let p∉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≤t≤b),n(γ,p)=θ(b)−θ(a)2π.

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

Facts & Assumptions

Given: A closed complex contour γ:[a,b]→C, a point p∉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)∈Z (The winding number of a closed contour is an integer), where n(γ,p)=(2πi)−1∫γdz/(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 2πiZ (Continuous logarithms and continuous arguments along a contour, Every contour missing a point admits a continuous logarithm, unique up to a constant in 2πiZ).

[L4]

For a complex contour γ, p∉γ∗ and a continuous logarithm λ of γ−p along γ, ∫γdz/(z−p)=λ(b)−λ(a) (The integral of dz/(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+bi 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)+iθ(t) as in [L6], the identity exp⁡(λ(t))=γ(t)−p of [L2] and the modulus formula [L3] give ∣γ(t)−p∣=eRe⁡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], 2πi n(γ,p)=λ(b)−λ(a), which step 2.1 rewrites as i(θ(b)−θ(a)); dividing by 2πi gives n(γ,p)=(θ(b)−θ(a))/(2π).

4.1step 3.1L1L2L8∎

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

Depends on

Used by

Dependency tree · two levels

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Sources