Math 411: Honours Complex Variables - University of Alberta

Define
g: C → C, z ↦→
∞�
n=0
(−1) n z 2n
(2n+1)! .
Then g is holomorphic and extends f1. Hence, f1 has a removable singularity at 0.
For m ≥ 2 and z �= 0, note that fm(z) = g(z)
zm−1. Since g(0) �= 0, we see that fm has
a pole of order m−1 at 0.
Example. Consider
f: C\{0} → C, z ↦→ e 1
z.
� �
1
Then 0 is not removable because lim f
n→∞ n
for f either: for n ∈ N, we have
�
�
�
�f � ��
i ���
= |e
n
−in |= 1.
55
= lim
n→∞ e n = ∞. But 0 is not a pole
Definition. Let D ⊂ C be open, let f: D → C be holomorphic, and let z0 ∈ C\D
be an isolated singularity of f. Then z0 is called essential if it is neither removable
nor a pole.
Theorem 8.3 (Casorati–Weierstraß Theorem). Let D ⊂ C be open, let f: D → C be
holomorphic, and let z0 ∈ C\D be an isolated singularity of f. Then z0 is essential
⇐⇒ f(Bǫ(z0)∩D) = C for each ǫ > 0.
Proof. “⇐” For each n ∈ N choose zn ∈ B 1
n
follows that lim
n→∞ |f(zn)|= ∞. Hence, z0 cannot be removable.
For each n ∈ N, choose z ′ n
∈ B 1
n
(z0) ∩ D such that |f(zn) − n|< 1.
It n
(z0)∩D such that |f(z ′ n
1 )|< . This means that
lim
n→∞ f(z′ n ) = 0, so that z0 is not a pole either.
“⇒” Assume for some ǫ0 > 0 that f(Bǫ0(z0)∩D) �= C. Without loss of generality,
suppose that Bǫ0(z0) \ {z0} ⊂ D. Let w0 ∈ C and δ > 0 be such that Bδ(w0) ⊂
C\f(Bǫ0(z0)\{z0}). Consider
Then g is holomorphic with
g: Bǫ0(z0)\{z0} → C, z ↦→
|g(z)|=
1
|f(z)−w0|
1
.
f(z)−w0
for z ∈ Bǫ0(z0)\{z0}. Hence, z0 is a removable singularity of g. Let ˜g: Bǫ0(z0) → C
be a holomorphic extension of g.
Case 1: ˜g(z0) �= 0. Since f(z) = 1
˜g(z) + w0 for z ∈ Bǫ0(z0) \ {z0}, this means
that z0 is a removable singularity of f, contradicting the fact that z0 is an essential
singularity.
≤ 1
δ
n