Math 411: Honours Complex Variables - University of Alberta

60 CHAPTER 9. HOLOMORPHIC FUNCTIONS ON ANNULI
Then F is entire with lim |F(z)|= lim |h2(z) − h1(z)|= 0. Hence, F is bounded
|z|→∞ |z|→∞
andentire and thus constant by Liouville’s theorem. Since lim
|z|→∞
|F(z)|= 0, this means
that F ≡ 0, so that g1 = g2 and h1 = h2.
To show that g and h exists, for z ∈ Ar,R(z0) choose ρ and P such that
Define
r < ρ < |z −z0|< P < R.
G: Ar,R(z0) → C, ζ ↦→
� f(ζ)−f(z)
ζ−z , ζ �= z,
f ′ (z), ζ = z.
ThenGiscertainlyholomorphiconAr,R(z0)\{z}. BecauseitiscontinuousonAr,R(z0),
Riemann’sRemovabilityCriterionimpliesthatGisinfactholomorphiconallofAr,R(z0).
It follows from Cauchy’s Integral Theorem for Annuli that
� �
G(ζ)dζ = G(ζ)dζ,
i.e.
�
∂Bρ(z0)
�
f(ζ)
dζ −f(z)
ζ −z
∂Bρ(z0)
∂Bρ(z0)
1
ζ −z dζ
� �� �
=0
�
=
∂BP(z0)
∂BP(z0)
Let us define the holomorphic functions (cf. Lemma 5.4)
h(z) := −1
�
f(ζ)
2πi ζ −z dζ
on C\Bρ[z0] and
∂Bρ(z0)
g(z) := 1
�
f(ζ)
2πi ∂BP(z0) ζ −z dζ
�
f(ζ) 1
dζ −f(z)
ζ −z ∂BP(z0) ζ −z dζ,
� �� �
=2πi
on BP(z0), noting from Cauchy’s Integral Theorem for Annuli that these definitions
are independent of the precise choices of ρ ∈ (r,|z−z0|) and P ∈ (|z−z0|,R). Then
the above result may be expressed as
and thus
Finally, we note that h satisfies
|h(z)|≤ ρ sup
−2πih(z) = 2πig(z)−2πif(z)
ζ∈∂Bρ(z0)
f(z) = g(z)+h(z).
� � sup |f(ζ)|
�
�
f(ζ) � ζ∈∂Bρ(z0)
�
�ζ
−z � ≤ ρ
dist(z,∂Bρ(z0)) −→
|z|→∞ 0.