Prove that if fz z iz 1 has a branch of log fz in domain

Prove that if f(z) = (z - i)/(z - 1) has a branch of log f(z) in domain D, then n(gamma, i) - n(gamma,1) = 0 for all closed p.w.s. paths gamma in D. (Do not simply quote a theorem.)

Solution

Let GCbe open and connected then there exist a branch of log(z) on G

if and only if: a. 0G   b. 1zdz=0

for every closed which is piecewise C1and s.t trace()G

I thought of denoting G=f(D(z0,))

and then we have some g s.t for every z(D(z0,)

e^g(f(z))=f(z) and by denoting h=gf

we get e^h(z)=f(z) and h is the required function.

0G - Since f does not vanish in D(z0,)

2. G is connected - since fH(D(z0,r))
Up to the continuity f(z) at z=z₀, there exists a number s. t. f(z) has no zeros in the disk D(z0,).

For zD(z0,) we define g(z):=[z₀,z]f(t)f(t)dt+logf(z₀),

where the integration is done over the interval [z₀,z] and the principal value of logf(z₀) is chosen. It is clear that gH(D(z₀,)), being an integral of an analytic function.

Next, the derivative of

(e^g(z)f(z))=e^g(z)g(z)f(z)+e^g(z)f(z)=e^g(z)(f(z)f(z)f(z)+f(z))=0 in D(z0,).

This implies e^g(z)f(z)=const there.

The substitution z=z₀ implies const=1 and f(z)=e^g(z).