712 CHAPTER 35. ANALYTIC FUNCTIONS

= limp→∞

12πi

p

∑k=0

∫γR

(z− z0)k

(w− z0)k+1 f (w)dw (35.7)

Which by definition is

∞

∑k=0

(1

2πi

∫γR

1

(w− z0)k+1 f (w)dw

)(z− z0)

k

It is assumed that f is continuous on Ui ∪ γ∗R. Thus there is an upper bound for | f (w)|,called M thanks to the extreme value theorem. Then for w ∈ γ∗R,∣∣∣∣∣ f (w)(z− z0)

k

(w− z0)k+1

∣∣∣∣∣≤ 1|z− z0|

M(|z− z0|

R

)k+1

<MR

(|z− z0|

R

)k

and |z− z0|/R < 1 so the right side is summable. Therefore, by Theorem 13.8.3, conver-gence is indeed uniform on γ∗R and so

f (z) =∞

∑k=0

(1

2πi

∫γR

1

(w− z0)k+1 f (w)dw

)(z− z0)

k ≡∞

∑k=0

ak (z− z0)k

This proves part of the next theorem which says, among other things, that when f hasone derivative on the interior of a disk, then it must have all derivatives.

Theorem 35.6.3 Suppose z0 ∈ U, an open set in C and f : U → X has a derivative foreach z ∈U. Then if B(z0,R)⊆U, then for each z ∈ B(z0,R) ,

f (z) =∞

∑n=0

an (z− z0)n . (35.8)

wherean ≡

12πi

∫γR

1

(w− z0)n+1 f (w)dw

and γR is a positively oriented parametrization for the circle bounding B(z0,R). Then

f (k) (z0) = k!ak, (35.9)

lim supn→∞

|an|1/n |z− z0|< 1, (35.10)

f (k) (z) =∞

∑n=k

n(n−1) · · ·(n− k+1)an (z− z0)n−k , (35.11)

Proof: 35.8 follows from the above argument. Now consider 35.10. The above argu-ment based on the Cauchy integral formula for a disk shows that if R > |ẑ− z0|> |z− z0| ,then

f (ẑ) =∞

∑n=0

an (ẑ− z0)n

and so, by the root test, Theorem 13.7.1,

1≥ lim supn→∞

|an|1/n |ẑ− z0|> lim supn→∞

|an|1/n |z− z0|