Math 411: Honours Complex Variables - University of Alberta

Proof. Let ζ ∈ ∂Br(z0), and note that
1
ζ −z =
1
(ζ −z0)−(z −z0)
= 1
ζ −z0
1
1− z−z0
ζ−z0
�
�
Since |z −z0|< r and |ζ −z0|= r, we have � z−z0
�
�
� < 1, so that
ζ−z0
1
ζ −z =
∞�
� �n 1 z −z0
=
(ζ −z0) ζ −z0 n=0
�
� (z−z0) n
Since � (ζ−z0) n+1
�
�
� =
lute and uniform convergence on ∂Br(z0).
∞�
n=0
1
(ζ −z0) n+1(z −z0) n .
|z−z0| n
rn+1 and �∞ |z−z0|
n=0
n
rn+1 < ∞, the Weierstraß M-test yields abso-
Theorem 6.3 (Power Series for Holomorphic Functions). Let D ⊂ C be open. Then
the following are equivalent for f: D → C:
(i) f is holomorphic;
(ii) for each z0 ∈ D, there exists r > 0 with Br(z0) ⊂ D and a0,a1,a2,... ∈ C such
that f(z) = � ∞
n=0 an(z −z0) n for all z ∈ Br(z0);
(iii) for each z0 ∈ D and r > 0 with Br(z0) ⊂ D, we have
for all z ∈ Br(z0).
f(z) =
∞�
n=0
f (n) (z0)
(z −z0)
n!
n
Proof. (iii) =⇒ (ii) is trivial; (ii) =⇒ (i) follows from Theorem 3.2.
(i) =⇒ (iii): Let z0 ∈ D, and let r > 0 be such that Br(z0) ⊂ D. Let z ∈ Br(z0)
and choose ρ ∈ (0,r) such that z ∈ Bρ(z0). Then we have:
f(z) = 1
�
f(ζ)
2πi ∂Bρ(z0) ζ −z dζ
= 1
� ∞� f(ζ)
2πi ∂Bρ(z0) (ζ −z0)
n=0
n+1(z −z0) n dζ, by Lemma 6.3,
∞�
� � �
1 f(ζ)
=
dζ (z −z0)
2πi
n=0 ∂Bρ(z0) (ζ −z0) n+1 n , by Lemma 6.1,
∞� f
=
(n) (z0)
(z −z0)
n!
n .
n=0
.
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