1790 CHAPTER 57. INFINITE PRODUCTS

because the function log(1− z) and the analytic function, −∑∞n=1

zn

n both are equal toln(1− x) on the real line segment (−1,1) , a set which has a limit point. Therefore, us-ing Lemma 57.1.1, ∣∣Ep (z)−1

∣∣=

∣∣∣∣(1− z)exp(

z+z2

2+ · · ·+ zp

p

)−1∣∣∣∣

=

∣∣∣∣∣(1− z)exp

(log((1− z)−1

)−

∞

∑n=p+1

zn

n

)−1

∣∣∣∣∣=

∣∣∣∣∣exp

(−

∞

∑n=p+1

zn

n

)−1

∣∣∣∣∣≤

∣∣∣∣∣− ∞

∑n=p+1

zn

n

∣∣∣∣∣e∣∣∣−∑

∞n=p+1

znn

∣∣∣

≤ 1p+1

·2 · e1/(p+1) |z|p+1 .≤ 3 |z|p+1

This proves the corollary.With this estimate, it is easy to prove the Weierstrass product formula.

Theorem 57.1.4 Let {zn} be a sequence of nonzero complex numbers which have no limitpoint in C and let {pn} be a sequence of nonnegative integers such that

∞

∑n=1

(R|zn|

)pn+1

< ∞ (57.1.5)

for all R ∈ R. Then

P(z)≡∞

∏n=1

Epn

(zzn

)is analytic on C and has a zero at each point, zn and at no others. If w occurs m times in{zn} , then P has a zero of order m at w.

Proof: Since {zn} has no limit point, it follows limn→∞ |zn| = ∞. Therefore, if pn =n−1 the condition, 57.1.5 holds for this choice of pn. Now by Theorem 57.0.2, the infiniteproduct in this theorem will converge uniformly on |z| ≤ R if the same is true of the sum,

∞

∑n=1

∣∣∣∣Epn

(zzn

)−1∣∣∣∣ . (57.1.6)

But by Corollary 57.1.3 the nth term of this sum satisfies∣∣∣∣Epn

(zzn

)−1∣∣∣∣≤ 3

∣∣∣∣ zzn

∣∣∣∣pn+1

.