Math 411: Honours Complex Variables - University of Alberta

90CHAPTER13. FUNCTIONTHEORETICCONSEQUENCESOFTHERESIDUETHEOREM
Proof. It is easily checked that the meromorphic functions do indeed form a commutative
ring. For each meromorphic function f �≡ 0 on D define
˜f: D → C, z ↦→ 1
f(z) .
AsP(f)isdiscrete, D\P(f)isconnectedbyLemma13.1. FromtheIdentityTheorem,
we then conclude that Z(f) is discrete, too. Thus ˜ f is meromorphic and (f ˜ f)(z) = 1
(the multiplicative identity) for z ∈ D.
Definition. Let z0 ∈ Z(f). If f(z) = (z−z0) k g(z), where g is a holomorphic function
with g(z0) �= 0, we say that k := ord(f,z0).
Theorem 13.2 (Argument Principle). Let D ⊂ C be open and simply connected, let
f be meromorphic on D, and let γ be a closed curve in D \(P(f)∪Z(f)). Then we
have �
1
2πi
γ
f ′ (ζ) �
dζ = ν(γ,z)ord(f,z)−
f(ζ)
z∈Z(f)
�
ν(γ,z)ord(f,z).
z∈P(f)
Proof. By the Residue Theorem, we have
�
1 f
2πi γ
′ (ζ) �
� ′ f
dζ = ν(γ,z) res
f(ζ) f
z∈Z(f)
,z
�
+ �
z∈P(f)
� ′ f
ν(γ,z) res
f ,z
�
.
Let z0 ∈ Z(f), and let k := ord(f,z). Then there is a holomorphic function g
with g(z0) �= 0 such that f(z) = (z −z0) k g(z) and thus
It follows that
for z near z0, so that
f ′ (z) = k(z −z0) k−1 g(z)+(z −z0) k g ′ (z).
f ′ (z)
f(z)
k
= +
z −z0
g′ (z)
g(z)
� ′ f
res
f ,z
�
= k.
Let z0 ∈ P(f), and let k := ord(f,z0). Then f(z) =
holomorphic such that g(z0) �= 0 and, consequently,
It follows that
for z �= z0 near z0, so that
f ′ (z) = −k(z −z0) −(k+1) g(z)+(z −z0) −k g ′ (z).
f ′ (z)
f(z)
−k
= +
z −z0
g′ (z)
g(z)
� ′ f
res
f ,z
�
= −k.
g(z)
holds with g
(z −z0) k