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The winding number vanishes on the unbounded component of the complement of the trace

Statement

Let γ be a closed complex contour with trace γ and length L(γ). Then Cγ has exactly one unbounded connected component U, and

n(γ,p)=0for every pU.

More precisely, if R>0 satisfies γ{z:zR} and p>R+L(γ)/(2π), then pU and n(γ,p)=0.

Facts & Assumptions

Given: A closed complex contour γ:[a,b]C.

[L1]

For a compact KC, the complement CK has exactly one unbounded connected component U, every other component is bounded, and {z:z>R}U whenever R>0 satisfies K{z:zR} (The complement of a compact plane set has exactly one unbounded connected component).

[L2]

The index n(γ,) is constant on every connected component of Cγ (The winding number is constant on each connected component of the complement of the trace).

[L3]

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).

[L4]

If f(z)M on the trace of a rectifiable contour γ, with M0, then γf(z)dzML(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L5]

n(γ,p)=(2πi)1γdz/(zp) (The winding number of a closed contour about a point off its trace).

[L7]

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

[L8]

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).

[L10]

The integers form an ordered commutative ring and their canonical image in R is discrete; in particular the only integer of modulus below 1 is 0 (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 mx<m+1).

Proof

technique · direct
1.1

The trace γ is the continuous image of a compact interval, hence compact by [L6], and bounded by [L6], so there is R>0 with γ{z:zR}. By [L1] the set Cγ has exactly one unbounded component U, and {z:z>R}U.

givenL1L6L8
2.1

Let p>R. For wγ one has wR, so wppwpR>0 by [L9]; hence pγ and 1/(zp)1/(pR) on the trace. By [L4] and [L5], n(γ,p)L(γ)/(2π(pR)).

step 1.1L4L5L9
3.1

If in addition p>R+L(γ)/(2π) then pR>L(γ)/(2π), so step 2.1 gives n(γ,p)<1; since n(γ,p) is an integer by [L3], it is 0 by [L10]. Such p exist, for instance p=R+L(γ)/(2π)+1, and each lies in U by step 1.1.

step 1.1step 2.1L3L10
4.1

By [L2] the index is constant on the connected component U, and step 3.1 exhibits a point of U where its value is 0; hence n(γ,p)=0 for every pU, which by [L7] contains every connected unbounded subset of Cγ that meets it.

step 3.1L2L7

Depends on

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