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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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A circle traversed k times has winding number k inside and 0 outside

Statement

Let a∈C, let r>0, let k∈Z and put

γk(t)=a+rexp⁡(ikt),t∈[0,2π].

Then γk is a closed complex contour and

n(γk,z)=kfor ∣z−a∣<r,n(γk,z)=0for ∣z−a∣>r.

For k≠0 the trace of γk 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∈C, a real r>0 and an integer k.

[L1]

For a closed complex contour γ and p∉γ∗, n(γ,p)=(2πi)−1∫γdz/(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∈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]→C, a point 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); 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+rexp⁡(it) with r>0, (2πi)−1∫σdz/(z−a)=1 (The normalized integral around a positively oriented circle centred at a is 1).

[L8]

exp⁡(z+w)=exp⁡zexp⁡w for all complex z,w, and exp⁡(x+0i)=ex for real x (exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential); for real x,y, exp⁡(x+iy)=ex(cos⁡y+isin⁡y) and ∣exp⁡(x+iy)∣=ex (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0).

[L10]

t↦(cos⁡t,sin⁡t) is a bijection from [0,2π) onto the unit circle S1 (t↦(cos⁡t,sin⁡t) is a bijection from [0,2π) onto the real unit circle), and cos⁡ and sin⁡ have fundamental period 2π (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-C1 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 Rm 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]

∣zw∣=∣z∣∣w∣ and ∣z+w∣≤∣z∣+∣w∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[L18]

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

Proof

technique · direct
1.1givenL8L9L11L12

By [L8] one has γk(t)=a+rcos⁡(kt)+irsin⁡(kt), which by [L11] is differentiable in t with the continuous derivative −krsin⁡(kt)+ikrcos⁡(kt), so γk is rectifiable by [L12]; and γk(2π)=a+rexp⁡(2πik)=a+r=γk(0) by [L9], so γk is a closed complex contour.

1.2givenL6L8L13L17

Since ∣γk(t)−a∣=r ∣exp⁡(ikt)∣=r>0 by [L8] and [L17], the centre a lies off the trace, and λ(t)=log⁡r+ikt is a continuous map on [0,2π] with exp⁡(λ(t))=elog⁡rexp⁡(ikt)=rexp⁡(ikt)=γk(t)−a by [L8] and [L13]; so λ is a continuous logarithm of γk−a along γk in the sense of [L6].

1.3givenL8L10L17

The trace of γk is {a+rexp⁡(ikt):0≤t≤2π}. If k=0 this is {a+r}. If k≠0 then {kt:0≤t≤2π} is a closed interval of length 2π∣k∣≥2π, so by the periodicity and surjectivity in [L10] the values exp⁡(iks) 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, ∫γkdz/(z−a)=λ(2π)−λ(0)=2πik, so n(γk,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⊆C∖γk∗ and a∈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∖γk∗, 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∖γk∗ containing D, and D is a connected subset of that complement containing a, so by [L15] it lies in one component; hence n(γk,z)=n(γk,a)=k for every z∈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(γk,z)=0 for every z∈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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