1682 CHAPTER 53. RESIDUES

Lemma 53.2.2 Let γr (t) ≡ a+ reit for t ∈ [0,2π] and let |z−a| < r. Then n(γr,z) = 1. If|z−a|> r, then n(γr,z) = 0.

Proof: For the first claim, consider for t ∈ [0,1] ,

f (t)≡ n(γr,a+ t (z−a)) .

Then from properties of the winding number derived earlier, f (t) ∈ Z, f is continuous,and f (0) = 1. Therefore, f (t) = 1 for all t ∈ [0,1] . This proves the first claim becausef (1) = n(γr,z) .

For the second claim,

n(γr,z) =1

2πi

∫γr

1w− z

dw

=1

2πi

∫γr

1w−a− (z−a)

dw

=1

2πi−1

z−a

∫γr

11−(w−a

z−a

)dw

=−1

2πi(z−a)

∫γr

∞

∑k=0

(w−az−a

)k

dw.

The series converges uniformly for w ∈ γr because∣∣∣∣w−az−a

∣∣∣∣= rr+ c

for some c > 0 due to the assumption that |z−a| > r. Therefore, the sum and the integralcan be interchanged to give

n(γr,z) =−1

2πi(z−a)

∞

∑k=0

∫γr

(w−az−a

)k

dw = 0

because w→(w−a

z−a

)k has an antiderivative. This proves the lemma.Now consider the following picture which pertains to the next lemma.

γr

a