Math 411: Honours Complex Variables - University of Alberta

54 CHAPTER 8. THE SINGULARITIES OF A HOLOMORPHIC FUNCTION
is holomorphic and bounded and thus, by Riemann’s Removability Criterion, has a
holomorphic extension h: Br(z0) → C with h(z0) = lim 1/f(z) = 0. Note that z0 is
z→z0
the only zero of h. Let
∞�
h(z) = an(z −z0) n
n=0
for z ∈ Br(z0) bethe power series representation of h. Set k := min{n ∈ N0 : an �= 0}.
Since a0 = h(z0) = 0, we have k ≥ 1. Define
˜h: Br(z0) → C, z ↦→
∞�
an(z −z0) n−k .
Then ˜ h is holomorphic, has no zeros, and satisfies h(z) = (z−z0) k˜ h(z) for z ∈ Br(z0).
For z ∈ Br(z0)\{z0}, we thus have
f(z) =
n=k
1
(z −z0) k˜ h(z) ,
so that we can construct the holomorphic function
�
(z −z0)
g: D∪{z0} C, z ↦→
kf(z), z �= z0,
1
˜h(z0)
, z = z0.
Definition. Let D ⊂ C be open, let f: D → C be holomorphic, and let z0 ∈ C\D
be a pole of f. Then the positive integer k in Theorem 8.2 is called the order of z0
and denoted by ord(f,z0). If ord(f,z0) = 1, we call z0 a simple pole of f.
Example. For m ∈ N, consider
fm: C\{0} → C, z ↦→ sinz
.
zm We claim that f1 has a removable singularity at 0 whereas fm has a pole of order
m−1 at 0 for m ≥ 2.
Recall that
sinz =
for z ∈ C. For z �= 0, we thus have
∞�
n=0
f1(z) = sinz
z =
(−1) n z 2n+1
(2n+1)!
∞�
n=0
(−1) n z 2n
(2n+1)! .