Math 411: Honours Complex Variables - University of Alberta

92CHAPTER13. FUNCTIONTHEORETICCONSEQUENCESOFTHERESIDUETHEOREM
Proof. Suppose that f is not constant. Let z0 ∈ D be arbitrary, and define
gn: D \{z0} → C, z ↦→ fn(z)−fn(z0)
for n ∈ N. Then g1,g2,... have no zeros. Since f is not constant, the function
D \{z0} → C, z ↦→ f(z)−f(z0)
is not zero, so that it has no zeros by Hurwitz’s theorem, i.e. f(z) �= f(z0) for all
z ∈ D, z �= z0.
Theorem 13.5 (Rouché’s Theorem). Let D ⊂ C be open and simply connected, and
let f,g : D → C be holomorphic. Suppose that γ is a closed curve in D such that
intγ = {z ∈ D \{γ} : ν(γ,z) = 1} and that
|f(ζ)−g(ζ)|< |f(ζ)|
for ζ ∈ {γ}. Then f and g have the same number of zeros in intγ (counting multiplicity).
Proof. For t ∈ [0,1], define ht := f +t(g−f), so that h0 = f and h1 = g. Also, since
|t(g −f)|≤ |g −f|< |f|
for any t ∈ [0,1] on {γ}, the functions ht have no zeros on {γ}. For t ∈ [0,1], let
n(t) ∈ N0 denote the number of zeros of ht in intγ. Since the functions ht have no
poles in D, we know from the Argument Principle that
n(t) = 1
2πi
�
γ
h ′ t(ζ) 1
dζ =
ht(ζ) 2πi
�
γ
f ′ (ζ)+t(g ′ (ζ)−f ′ (ζ))
f(ζ)+t(g(ζ)−f(ζ)) dζ.
We thus see that n(t) is a continuous function of t. But since n(t) can only taken on
integer values, it must be constant on [0,1]; in particular, n(0) = n(1).
Example. How many zeros does z 4 −4z +2 have in D?
Set
g(z) := z 4 −4z +2 and f(z) = −4z +2.
For ζ ∈ ∂D, we have |f(ζ)|≥ |−4z|−2 = 4−2 = 2, so that
|f(ζ)−g(ζ)|= |ζ 4 |= 1 < 2 ≤ |f(ζ)|.
Since f has precisely one zero in D, so does g.
Corollary 13.5.1 (Fundamental Theorem of Algebra). Let p be a polynomial with
n := degp ≥ 1. Then p has n zeros (counting multiplicity).