1644 CHAPTER 51. FUNDAMENTALS OF COMPLEX ANALYSIS

because z ∈ H. This shows g(z) is well defined. Also, g is analytic on Ω because it equalsh there. It is routine to verify that g is analytic on H also because of the second line of51.7.27. (See discussion at the end if this is not clear. )

Therefore, g is an entire function, meaning that it is analytic on all of C.Now note that ∑

mk=1 n(γk,z) = 0 for all z contained in the unbounded component of

C\ ∪mk=1 γ∗k which component contains B(0,r)C for r large enough. It follows that for

|z| > r, it must be the case that z ∈ H and so for such z, the bottom description of g(z)found in 51.7.27 is valid. Therefore, it follows

lim|z|→∞

∥g(z)∥= 0

and so g is bounded and analytic on all of C. By Liouville’s theorem, g is a constant.Hence, from the above equation, the constant can only equal zero.

For z ∈Ω\∪mk=1γ∗k , since it was just shown that h(z) = g(z) = 0 on Ω

0 = h(z) =1

2πi

m

∑k=1

∫γk

φ (z,w)dw =1

2πi

m

∑k=1

∫γk

f (w)− f (z)w− z

dw =

12πi

m

∑k=1

∫γk

f (w)w− z

dw− f (z)m

∑k=1

n(γk,z) .

In case it is not obvious why g is analytic on H, use the formula. It reduces to showingthat

z→∫

γk

f (w)w− z

dw

is analytic. However, taking a difference quotient and simplifying a little, one obtains∫γk

f (w)w−(z+h)dw−

∫γk

f (w)w−z dw

h=∫

γk

f (w)(w− z)(w− (z+h))

dw

considering only small h, the denominator is bounded below by some δ > 0 and also f (w)is bounded on the compact set γ∗k , | f (w)| ≤M. Then for such small h,∣∣∣∣∣ f (w)

(w− z)(w− (z+h))− f (w)

(w− z)2

∣∣∣∣∣=

∣∣∣∣ 1w− z

(1

(w− (z+h))− 1

(w− z)

)f (w)

∣∣∣∣≤

∣∣∣∣ 1w− z

∣∣∣∣ 1δ

hM

it follows that one obtains uniform convergence as h→ 0 of the integrand to f (w)(w−z)2 for any

sequence h→ 0 and so the integral converges to∫γk

f (w)

(w− z)2 dw