Math 411: Honours Complex Variables - University of Alberta

Chapter 10
The Winding Number of a Curve
Definition. Let γ be a closed curve in C, and let z ∈ C \ {γ}. Then the winding
number of γ with respect to z is defined as
ν(γ,z) := 1
�
1
2πi ζ −z dζ.
Remark. Geometrically, ν(γ,z) is the number of times γ winds around z in the counterclockwise
direction.
Lemma 10.1. Let γ: [0,1] → C be a curve, and let z ∈ C \{γ}. Then there exist
open discs D1,...,Dn ⊂ C\{z} and a partition 0 = t0 < t1 < ··· < tn = 1 such that
γ([tj−1,tj]) ⊂ Dj for j = 1,...,n.
Proof. Let ǫ := dist(z,{γ}) > 0. Since γ is uniformly continuous, there exists δ > 0
such that |γ(t)−γ(t ′ )|< ǫ for all t,t ′ ∈ [0,1] such that |t−t ′ |< δ. Choose 0 = t0 <
t1 < ··· < tn = 1 such that |tj−1−tj|< δ for j = 1,...,n, and set Dj := Bǫ(γ(tj)) for
j = 1,...,n. Bythechoice ofǫ, itisclearthatD1,...,Dn ⊂ C\{z}. Forj = 1,...,n,
let t ∈ [tj−1,tj], and note that |t−tj|≤ |tj−1 −tj|< δ, so that |γ(t)−γ(tj)|< ǫ, i.e.
γ(t) ∈ Dj; consequently, γ([tj−1,tj]) ⊂ Dj holds.
Proposition 10.1. Let γ be a closed curve in C, and let z ∈ C \ {γ}. Then
ν(γ,z) ∈ Z.
Proof. Choose a partition 0 = t0 < t1 < ··· < tn = 1 and open discs D1,...,Dn such
that γ([tj−1,tj]) ⊂ Dj for j = 1,...,n.
Let j ∈ {1,...,n}. Since z /∈ Dj, there exists a holomorphic function Lj: Dj → C
such that
e Lj(w) = w−z for w ∈ Dj.
On noting that γ(tn) = γ(t0), it is convenient to denote Dn+1 := D1 and Ln+1 := L1.
For j = 1,...,n we then see that γ(tj) ∈ Dj ∩Dj+1 and hence
exp(Lj(γ(tj))) = γ(tj)−z = exp(Lj+1(γ(tj))),
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γ