36.2. FUNCTIONS ANALYTIC ON AN ANNULUS 731

γR

γ̂r • z

Γ1Γ2 •z0

It follows from Theorem 35.7.1, that for z in the annulus,

12πi

∫γR

f (w)w− z

dw+1

2πi

∫γ̂r

f (w)w− z

dw = f (z)

This is because the contributions to the line integrals along those straight lines is 0 sincethey cancel off because of opposite orientations. Let γr be the opposite orientation from γ̂r.Then this reduces to ∫

γR

f (w)w− z

dw−∫

γr

f (w)w− z

dw = 2πi f (z)

Thus

f (z) =1

2πi

[∫γR

f (w)w− z0− (z− z0)

dw+∫

γr

f (w)(z− z0)− (w− z0)

dw]

=1

2πi

[∫γR

1w− z0

f (w)1− z−z0

w−z0

dw+∫

γr

1z− z0

f (w)1− w−z0

z−z0

dw

]

Now note that for z in the annulus between the two circles and w ∈ γ∗R,∣∣∣ z−z0

w−z0

∣∣∣< 1, and for

w ∈ γ∗r ,∣∣∣w−z0

z−z0

∣∣∣< 1. In fact, in each case, there is b < 1 such that

w ∈ γ∗R,

∣∣∣∣ z− z0

w− z0

∣∣∣∣< b < 1, w ∈ γ∗r ,

∣∣∣∣w− z0

z− z0

∣∣∣∣< b < 1 (36.3)

Thus you can use the formula for the sum of an infinite geometric series and conclude

f (z) =1

2πi

 ∫γR

f (w) 1w−z0

∑∞n=0

(z−z0w−z0

)ndw

+∫

γrf (w) 1

(z−z0)∑

∞n=0

(w−z0z−z0

)ndw

Then from the uniform estimates of 36.3, one can conclude uniform convergence of thepartial sums for w ∈ γ∗R or γ∗r , and so by the Weierstrass M test, Theorem 13.8.3, one can