1788 CHAPTER 57. INFINITE PRODUCTS∣∣∣∣∣ M

∏k=1

(1+uk (z0))−∞

∏k=1

(1+uk (z0))

∣∣∣∣∣≤

∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣∣∣∣∣∣1− ∞

∏k=M+1

(1+uk (z0))

∣∣∣∣∣≤

∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣∣∣∣∣∣ ∞

∏k=M+1

(1+ |uk (z0)|)−1

∣∣∣∣∣≤ e

∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣∣∣∣∣∣ln ∞

∏k=M+1

(1+ |uk (z0)|)− ln1

∣∣∣∣∣≤ e

(∞

∑k=M+1

ln(1+ |uk (z)|)

)∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣≤ e

∞

∑k=M+1

|uk (z)|

∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣≤ 1

2

∣∣∣∣∣ M

∏k=1

(1+uk (z0))

∣∣∣∣∣whenever M is large enough. Therefore, for such M,

M

∏k=1

(1+uk (z0)) = 0

and so uk (z0) =−1 for some k ≤M. This proves the theorem.

57.1 Analytic Function With Prescribed ZerosSuppose you are given complex numbers, {zn} and you want to find an analytic func-tion, f such that these numbers are the zeros of f . How can you do it? The prob-lem is easy if there are only finitely many of these zeros, {z1,z2, · · · ,zm} . You just write(z− z1)(z− z2) · · ·(z− zm) . Now if none of the zk = 0 you could also write it as

m

∏k=1

(1− z

zk

)and this might have a better chance of success in the case of infinitely many prescribedzeros. However, you would need to verify something like ∑

∞n=1

∣∣∣ zzn

∣∣∣ < ∞ which might not

be so. The way around this is to adjust the product, making it ∏∞k=1

(1− z

zk

)egk(z) where

gk (z) is some analytic function. Recall also that for |x| < 1, ln((1− x)−1

)= ∑

∞n=1

xn

n . If

you had x/xn small and real, then 1 = (1− x/xn)exp(

ln((1− x/xn)

−1))

and ∏∞k=1 1 of

course converges but loses all the information about zeros. However, this is why it is not