1802 CHAPTER 57. INFINITE PRODUCTS

Now suppose f is analytic on B(0,r+ ε) , and f has no zeros on B(0,r). Then you candefine a branch of the logarithm which makes sense for complex numbers near f (z) . Thusz→ log( f (z)) is analytic on B(0,r+ ε). Therefore, its real part, u(x,y) ≡ ln | f (x+ iy)|must be harmonic. Consider the following lemma.

Lemma 57.4.2 Let u be harmonic on B(0,r+ ε) . Then

u(0) =1

2π

∫π

−π

u(

reiθ)

dθ .

Proof: For a harmonic function, u defined on B(0,r+ ε) , there exists an analytic func-tion, h = u+ iv where

v(x,y)≡∫ y

0ux (x, t)dt−

∫ x

0uy (t,0)dt.

By the Cauchy integral theorem,

h(0) =1

2πi

∫γr

h(z)z

dz =1

2π

∫π

−π

h(

reiθ)

dθ .

Therefore, considering the real part of h,

u(0) =1

2π

∫π

−π

u(

reiθ)

dθ .

This proves the lemma.Now this shows the following corollary.

Corollary 57.4.3 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

)∣∣∣ (57.4.21)

What if f has some zeros on |z| = r but none on B(0,r)? It turns out 57.4.21 is stillvalid. Suppose the zeros are at

{reiθ k

}mk=1 , listed according to multiplicity. Then let

g(z) =f (z)

∏mk=1 (z− reiθ k)

.

It follows g is analytic on B(0,r+ ε) but has no zeros in B(0,r). Then 57.4.21 holds for gin place of f . Thus

ln | f (0)|−m

∑k=1

ln |r|

=1

2π

∫π

−π

ln∣∣∣ f (reiθ

)∣∣∣dθ − 12π

∫π

−π

m

∑k=1

ln∣∣∣reiθ − reiθ k

∣∣∣dθ

=1

2π

∫π

−π

ln∣∣∣ f (reiθ

)∣∣∣dθ − 12π

∫π

−π

m

∑k=1

ln∣∣∣eiθ − eiθ k

∣∣∣dθ −m

∑k=1

ln |r|

=1

2π

∫π

−π

ln∣∣∣ f (reiθ

)∣∣∣dθ − 12π

∫π

−π

m

∑k=1

ln∣∣∣eiθ −1

∣∣∣dθ −m

∑k=1

ln |r|