358 CHAPTER 14. FUNDAMENTALS

and now take the limit as r̂→ r and use uniform continuity of f on B(z0,r) to obtain f (z) =1

2πi∫

γr

f (w)w−z dw. ■

This is the Cauchy integral formula for a disk. This remarkable formula is sufficient toshow that if a function has a derivative, then it has infinitely many and in addition to this,the function can be represented as a power series. Let z0 be the center of the large circle.

In the situation of Theorem 14.8.1,

f (z) =1

2πi

∫γR

f (w)w− z0− (z− z0)

dw =1

2πi

∫γR

1w− z0

f (w)1− z−z0

w−z0

dw

Now∣∣∣ z−z0

w−z0

∣∣∣= |z−z0|R < 1 for all w ∈ γ∗R. Therefore, the above equals

12πi

∫γR

∞

∑k=0

f (w)(z− z0)k

(w− z0)k+1 dw =

12πi

∫γR

(∞

∑k=0

(z− z0)k

(w− z0)k+1

)f (w)dw

Since f is continuous, one can apply the Weierstrass M test Theorem 2.5.42 to concludethat the above series converges uniformly on γ∗R. Then the above reduces to

f (z) =1

2πi

∞

∑k=0

(∫γR

f (w)

(w− z0)k+1 dw

)(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 circle, then it must have all derivatives. Note that thefunction has values in a complex Banach space.

Theorem 14.8.3 Suppose z0 ∈U, an open set in C and f : U → X has a derivativefor each z ∈U. Then if B(z0,R)⊆U, then for each z ∈ B(z0,R) ,

f (z) =∞

∑n=0

an (z− z0)n . (14.18)

where an ≡ 12πi∫

γR1

(w−z0)n+1 f (w)dw ∈ X and γR is a positively oriented parametrization

for the circle bounding B(z0,R). Then

f (k) (z0) = k!ak, (14.19)

lim supn→∞

∥an∥1/n |z− z0|< 1, (14.20)

f (k) (z) =∞

∑n=k

n(n−1) · · ·(n− k+1)an (z− z0)n−k , (14.21)

Proof: 14.18 follows from the above argument. Now consider 14.20. 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 1.12.1,

1≥ lim supn→∞

∥an∥1/n |ẑ− z0|> lim supn→∞

∥an∥1/n |z− z0|