I have calculated $f^{-1}[B_2]$ the set with $x\in\mathbb{R}$ and $-\pi<Im(z)<\pi$. But in the solutions it says that it's $\pi/2$. Why is that?

@johnny09 You are asking when is the real part of $e^{a+bi} = e^a e^{bi} = e^a (\cos b+i \sin b) $ positive.

This is the same as asking when is $\cos b$ positive. You will get $-\pi/2 < b < \pi/2$.

Maybe it will help if you look at this picture showing the meaning of $e^{ib}$. Taken from the Wikipedia article on Euler's formula: en.wikipedia.org/wiki/Euler%27s_formula

Alright. Thanks! :)