15.2. FUNCTIONS ANALYTIC ON AN ANNULUS 377

r, and obtain the above formula valid for γ R̂ and γ r̂.Then pass to a limit using continuity off on ann(z0,r,R) to replace R̂ and r̂ with R and r. The details are routine and left for you.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 (15.4)

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 15.4 and the Weierstrass M test, Theorem 2.5.42, onecan conclude uniform convergence of the partial sums for w ∈ γ∗R or γ∗r and so one caninterchange the summation with the integral and write

f (z) =∞

∑n=0

(1

2πi

∫γR

f (w)1

(w− z0)n+1 dw

)(z− z0)

n

+∞

∑n=0

(1

2πi

∫γr

f (w)(w− z0)n dw

)1

(z− z0)n+1 ,

both series converging absolutely. Thus there are an,bn ∈ X such that

f (z) =∞

∑n=0

an (z− z0)n +

∞

∑n=1

bn (z− z0)−n

This proves the following theorem.

Theorem 15.2.2 Let z ∈ ann(z0,r,R) and let f : ann(z0,r,R)→ X be analytic onann(z0,r,R) and continuous on ann(z0,r,R). Then for any z ∈ ann(z0,r,R) ,

f (z) =∞

∑n=0

an (z− z0)n +

∞

∑n=1

bn (z− z0)−n (15.5)

where

an =1

2πi

∫γR

f (w)1

(w− z0)n+1 dw, bn =

12πi

∫γr

f (w)(w− z0)n−1 dw

and both of these series in 15.5 converge absolutely. If r < r̂ < R̂ < R, then convergence ofboth series is absolute and uniform for z ∈ ann

(z0, r̂, R̂

).