57.1. ANALYTIC FUNCTION WITH PRESCRIBED ZEROS 1789

too unreasonable to consider factors of the form(1− z

zk

)e∑

pkk=1

(z

zk

)k 1k

where pk is suitably chosen.First here are some estimates.

Lemma 57.1.1 For z ∈ C,|ez−1| ≤ |z|e|z|, (57.1.3)

and if |z| ≤ 1/2, ∣∣∣∣∣ ∞

∑k=m

zk

k

∣∣∣∣∣≤ 1m|z|m

1−|z|≤ 2

m|z|m ≤ 1

m1

2m−1 . (57.1.4)

Proof: Consider 57.1.3.

|ez−1|=

∣∣∣∣∣ ∞

∑k=1

zk

k!

∣∣∣∣∣≤ ∞

∑k=1

|z|k

k!= e|z|−1≤ |z|e|z|

the last inequality holding by the mean value theorem. Now consider 57.1.4.∣∣∣∣∣ ∞

∑k=m

zk

k

∣∣∣∣∣ ≤ ∞

∑k=m

|z|k

k≤ 1

m

∞

∑k=m|z|k

=1m|z|m

1−|z|≤ 2

m|z|m ≤ 1

m1

2m−1 .

This proves the lemma.The functions, Ep in the next definition are called the elementary factors.

Definition 57.1.2 Let E0 (z)≡ 1− z and for p≥ 1,

Ep (z)≡ (1− z)exp(

z+z2

2+ · · ·+ zp

p

)In terms of this new symbol, here is another estimate. A sharper inequality is available

in Rudin [113] but it is more difficult to obtain.

Corollary 57.1.3 For Ep defined above and |z| ≤ 1/2,∣∣Ep (z)−1∣∣≤ 3 |z|p+1 .

Proof: From elementary calculus, ln(1− x) =−∑∞n=1

xn

n for all |x|< 1. Therefore, for|z|< 1,

log(1− z) =−∞

∑n=1

zn

n, log

((1− z)−1

)=

∞

∑n=1

zn

n,