= Solution
Termwise <differentiation> inside the disc gives
$$
f'(z)=\sum_{n=1}^{\infty}(1-z)^{n-1}=\frac1z.
$$
Consequently
$$
\frac d{dz}\bigl(ze^{-f(z)})
=e^{-f(z)}(1-zf'(z))=0.
$$
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 $a=|a|e^{i\theta_a}$ with $0<\theta_a<2\pi$. On $|z/a-1|<1$, define
$$
\ell_a(z)=f(z/a)+\log|a|+i\theta_a.
$$
Then $e^{\ell_a(z)}=z$ and $\operatorname{Im}\ell_a(a)=\theta_a$. After shrinking the neighbourhood of $a$, continuity keeps its imaginary part in $(0,2\pi)$.
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