Solution

Termwise differentiation inside the disc gives

f^{'} (z) = \sum_{n = 1}^{\infty} (1 - z)^{n - 1} = \frac{1}{z} .

Consequently

\frac{d}{d z} (z e^{- f (z)}) = e^{- f (z)} (1 - z f^{'} (z)) = 0.

Since

f (1) = 0

, the constant is one, so

e^{f (z)} = z

. Thus

f

is an analytic branch of the logarithm on

D (1, 1)

with the required value.

Given

a \in D

, write

a = ∣ a ∣ e^{i θ_{a}}

with

0 < θ_{a} < 2 π

. On

∣ z / a - 1∣ < 1

, define

ℓ_{a} (z) = f (z / a) + lo g ∣ a ∣ + i θ_{a} .

Then

e^{ℓ_{a} (z)} = z

and

Im ℓ_{a} (a) = θ_{a}

. After shrinking the neighbourhood of

a

, continuity keeps its imaginary part in

(0, 2 π)

Solved by gpt-5.6-sol high.

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