1626 CHAPTER 51. FUNDAMENTALS OF COMPLEX ANALYSIS

Theorem 51.3.15 If f : Ω→ X is analytic and if B(z0,r)⊆Ω, then

f (z) =∞

∑n=0

an (z− z0)n (51.3.13)

for all |z− z0|< r. Furthermore,

an =f (n) (z0)

n!. (51.3.14)

Proof: Consider |z− z0|< r and let γ (t)= z0+reit , t ∈ [0,2π] . Then for w∈ γ ([0,2π]) ,∣∣∣∣ z− z0

w− z0

∣∣∣∣< 1

and so, by the Cauchy integral formula,

f (z) =1

2πi

∫γ

f (w)w− z

dw

=1

2πi

∫γ

f (w)

(w− z0)(

1− z−z0w−z0

)dw

=1

2πi

∫γ

f (w)(w− z0)

∞

∑n=0

(z− z0

w− z0

)n

dw.

Since the series converges uniformly, you can interchange the integral and the sum to obtain

f (z) =∞

∑n=0

(1

2πi

∫γ

f (w)

(w− z0)n+1

)(z− z0)

n

≡∞

∑n=0

an (z− z0)n

By Theorem 51.3.11, 51.3.14 holds.Note that this also implies that if a function is analytic on an open set, then all of its

derivatives are also analytic. This follows from Theorem 51.1.4 which says that a functiongiven by a power series has all derivatives on the disk of convergence.

51.4 Exercises1. Show that if |ek| ≤ ε, then

∣∣∑∞k=m ek

(rk− rk+1

)∣∣< ε if 0≤ r < 1. Hint: Let |θ |= 1and verify that

θ

∞

∑k=m

ek

(rk− rk+1

)=

∣∣∣∣∣ ∞

∑k=m

ek

(rk− rk+1

)∣∣∣∣∣= ∞

∑k=m

Re(θek)(

rk− rk+1)

where −ε < Re(θek)< ε.