Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
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Define<br />
⎧<br />
⎪⎨<br />
g: D ∪{z0} → C, z ↦→<br />
⎪⎩<br />
∞�<br />
an+2(z −z0) n , z ∈ Bǫ(z0),<br />
n=0<br />
Then g is a holomorphic function extending f.<br />
f(z), z ∈ D \Bǫ(z0).<br />
Definition. Let D ⊂ C be open, let f: D → C be holomorphic, and let z0 ∈ C\D<br />
be an isolated singularity <strong>of</strong> f. Then z0 is called a pole <strong>of</strong> f if lim|f(z)|=<br />
∞.<br />
z→z0<br />
Example. For n ∈ N, the function<br />
has a pole at 0.<br />
C\{0} → C, z ↦→ 1<br />
z n<br />
Theorem 8.2 (Poles). Let D ⊂ C be open, let f : D → C be holomorphic, and let<br />
z0 ∈ C\D be an isolated singularity <strong>of</strong> f. Then z0 is a pole <strong>of</strong> f ⇐⇒ there exist a<br />
unique k ∈ N and a holomorphic function g: D ∪{z0} → C such that g(z0) �= 0 and<br />
for z ∈ D.<br />
f(z) = g(z)<br />
(z −z0) k<br />
Pro<strong>of</strong>.<br />
”⇐” This follows directly from the definition <strong>of</strong> a pole.<br />
”⇒” Let us prove the uniqueness first. Suppose that there exist natural numbers<br />
k1 ≤ k2 and holomorphic functions g1,g2: D ∪{z0} → C such that gj(z0) �= 0 and<br />
f(z) = gj(z)<br />
(z −z0) kj<br />
for z ∈ D and j = 1,2. If k2 > k1, we then find for all z ∈ D that<br />
g2(z0) = lim<br />
z→z0<br />
g2(z) = lim(z<br />
−z0)<br />
z→z0<br />
k2−k1g1(z) = 0·g1(z0) = 0,<br />
which is a contradiction. Hence k1 = k2 and thus g1 = g2 on D and, by continuity,<br />
on D ∪{z0}.<br />
To establish the existence <strong>of</strong> k and g, choose r > 0 such that Br(z0)\{z0} ⊂ D<br />
and |f(z)|≥ 1 for all z ∈ Br(z0)\{z0}. Then<br />
Br(z0)\{z0} → C, z ↦→ 1<br />
f(z)<br />
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