57.4. JENSEN’S FORMULA 1803
Therefore, 57.4.21 will continue to hold exactly when 12π
∫π
−π ∑mk=1 ln
∣∣eiθ −1∣∣dθ = 0. But
this is the content of Lemma 57.4.1. This proves the following lemma.
Lemma 57.4.4 Suppose f is analytic on B(0,r+ ε) and has no zeros on B(0,r) . Then
ln | f (0)|= 12π
∫π
−π
ln∣∣∣ f (reiθ
)∣∣∣ (57.4.22)
With this preparation, it is now not too hard to prove Jensen’s formula. Suppose thereare n zeros of f in B(0,r) ,{ak}n
k=1, listed according to multiplicity, none equal to zero. Let
F (z)≡ f (z)n
∏i=1
r2−aizr (z−ai)
.
Then F is analytic on B(0,r+ ε) and has no zeros in B(0,r) . The reason for this is thatf (z)/∏
ni=1 r (z−ai) has no zeros there and r2−aiz cannot equal zero if |z|< r because if
this expression equals zero, then
|z|= r2
|ai|> r.
The other interesting thing about F (z) is that when z = reiθ ,
F(
reiθ)
= f(
reiθ) n
∏i=1
r2−aireiθ
r (reiθ −ai)
= f(
reiθ) n
∏i=1
r−aieiθ
(reiθ −ai)= f
(reiθ)
eiθn
∏i=1
re−iθ −ai
reiθ −ai
so∣∣F (reiθ
)∣∣= ∣∣ f (reiθ)∣∣.
Theorem 57.4.5 Let f be analytic on B(0,r+ ε) and suppose f (0) ̸= 0. If the zeros of fin B(0,r) are {ak}n
k=1, listed according to multiplicity, then
ln | f (0)|=−n
∑i=1
ln(
r|ai|
)+
12π
∫ 2π
0ln∣∣∣ f (reiθ
)∣∣∣dθ .
Proof: From the above discussion and Lemma 57.4.4,
ln |F (0)|= 12π
∫π
−π
ln∣∣∣ f (reiθ
)∣∣∣dθ
But F (0) = f (0)∏ni=1
rai
and so ln |F (0)|= ln | f (0)|+∑ni=1 ln
∣∣∣ rai
∣∣∣ . Therefore,
ln | f (0)|=−n
∑i=1
ln∣∣∣∣ rai
∣∣∣∣+ 12π
∫ 2π
0ln∣∣∣ f (reiθ
)∣∣∣dθ
as claimed.Written in terms of exponentials this is
| f (0)|n
∏k=1
∣∣∣∣ rak
∣∣∣∣= exp(
12π
∫ 2π
0ln∣∣∣ f (reiθ
)∣∣∣dθ
).