1792 CHAPTER 57. INFINITE PRODUCTS
Proof: There is nothing to prove if {zn} is finite. You just let f (z) = ∏mj=1 (z− z j)
where {zn}= {z1, · · · ,zm}.Pick w ∈ Ω \ {zn}∞
n=1 and let h(z) ≡ 1z−w . Since w is not a limit point of {zn} , there
exists r > 0 such that B(w,r) contains no points of {zn} . Let Ω1 ≡ Ω \ {w}. Now h isnot constant and so h(Ω1) is an open set by the open mapping theorem. In fact, h mapseach component of Ω to a region. |zn−w| > r for all zn and so |h(zn)| < r−1. Thus thesequence, {h(zn)} is a bounded sequence in the open set h(Ω1) . It has no limit point inh(Ω1) because this is true of {zn} and Ω1. By Lemma 57.1.6 there exist wn ∈ ∂ (h(Ω1))such that limn→∞ |wn−h(zn)|= 0. Consider for z ∈Ω1
f (z)≡∞
∏n=1
En
(h(zn)−wn
h(z)−wn
). (57.1.8)
Letting K be a compact subset of Ω1, h(K) is a compact subset of h(Ω1) and so if z ∈ K,then |h(z)−wn| is bounded below by a positive constant. Therefore, there exists N largeenough that for all z ∈ K and n≥ N,∣∣∣∣h(zn)−wn
h(z)−wn
∣∣∣∣< 12
and so by Corollary 57.1.3, for all z ∈ K and n≥ N,∣∣∣∣En
(h(zn)−wn
h(z)−wn
)−1∣∣∣∣≤ 3
(12
)n
. (57.1.9)
Therefore,∞
∑n=1
∣∣∣∣En
(h(zn)−wn
h(z)−wn
)−1∣∣∣∣
converges uniformly for z ∈ K. This implies ∏∞n=1 En
(h(zn)−wnh(z)−wn
)also converges uniformly
for z∈ K by Theorem 57.0.2. Since K is arbitrary, this shows f defined in 57.1.8 is analyticon Ω1.
Also if zn is listed m times so it is a zero of multiplicity m and wn is the point from∂ (h(Ω1)) closest to h(zn) , then there are m factors in 57.1.8 which are of the form
En
(h(zn)−wn
h(z)−wn
)=
(1− h(zn)−wn
h(z)−wn
)egn(z)
=
(h(z)−h(zn)
h(z)−wn
)egn(z)
=zn− z
(z−w)(zn−w)
(1
h(z)−wn
)egn(z)
= (z− zn)Gn (z) (57.1.10)
where Gn is an analytic function which is not zero at and near zn. Therefore, f has a zeroof order m at zn. This proves the theorem except for the point, w which has been left out