had a test a few days ago and was not sure about my solution. Please check it out and tell me if I was wrong, none I asked knew what to do \ others did like me.
EDIT: ( I really hope I am not lying about $f(2i)$, I 99% remember that was the function)
It could be that instead of $i\cdot \left(\frac{9}{2}\pi \right)$ it was: $\left(\frac{9}{2}\pi \right)$, but one of them for sure.
Let $log(z)$ be an analytic branch of $f$.
let $f$ be defined at the domain: $\mathbb{C}\backslash _{R_{\le 0}}$
we know that $f\left(2i\right)=ln(2)+i\cdot \left(\frac{9}{2}\pi \right)$
find the image of $f$
My solution: ( atleast what I tried )
I wrote $f(z)$ as:
$f(z)=ln(R)+i\cdot (\theta + 2\pi k)$, when $k\in\mathbb{Z}$
from here I can understand that k has to be 2, so $\theta$ has to be in the range of $f(2i)$.
which is $3\pi \le \theta <5\pi$
Thus receiving the conclusion of A=$Im(f)$ as of:
$A =${$ \left|z\right|=2,3\pi \le \theta <5\pi$}