Math 411: Honours Complex Variables - University of Alberta

62 CHAPTER 9. HOLOMORPHIC FUNCTIONS ON ANNULI
Finally, pick m ∈ Z and ρ ∈ (r,R). Note that
f(z)
=
(z −z0) m+1
∞�
n=−∞
an(z −z0) n−m−1 =
converges uniformly on ∂Bρ(z0). Hence, we find
�
∂Bρ(z0)
f(ζ)
dζ =
(ζ −z0) m+1
∞�
n=−∞
an+m+1
�
∂Bρ(z0)
∞�
n=−∞
(ζ−z0) n dζ = am
noting that (ζ −z0) n has an antiderivative for all n �= −1. Thus
am = 1
�
f(ζ)
dζ
2πi ∂Bρ(z0) (ζ −z0) m+1
an+m+1(z −z0) n
�
∂Bρ(z0)
1
dζ = 2πiam,
ζ −z0
Corollary 9.3.1. Let z0 ∈ C, let r > 0, and let f: Br(z0)\{z0} → C be holomorphic
with Laurent representation f(z) = � ∞
n=−∞ an(z−z0) n . Then the singularity z0 of f
is
(i) removable if and only if an = 0 for n < 0;
(ii) a pole of order k ∈ N if and only if a−k �= 0 and an = 0 for all n < −k;
(iii) essential if and only if an �= 0 for infinitely many n < 0.
Proof.
(i) The “if” part follows from Theorem 6.3.
Conversely, suppose that z0 is a removable singularity, and let ˜ f : Br(z0) → C
be a holomorphic extension of f with Taylor expansion ˜ f(z) = �∞ n
n=0bn(z−z0) for z ∈ Br(z0). The uniqueness of the Laurent representation yields an = bn for
n ∈ N0 and an = 0 for n < 0.
(ii) For the “if” part, set
g(z) := (z −z0) k f(z) =
∞�
n=−k
an(z −z0) n+k
for z ∈ Br(z0)\{z0}. Then g extends holomorphically to Br(z0) with g(z0) =
a−k �= 0. By definition, we have f(z) = g(z)
(z−z0) k for z ∈ Br(z0)\{z0}. Hence, f
has a pole of order k at z0.