If z=a+bi then f(z)=f(a+bi)=$e^{a+bi}=e^{a}e^{bi}$. For each of the following complex number $w_{i}$ find a $z_{i}$ so that $f(z_{i})=w_{i}$. If no $z_{i}$ exists explain why not.
a) $w_{1}=e^{2}$
b) $w_{2}=0$
c) $w_{3}=-3i$
So I was able to compute part a) and part c), but I'm confused about how to explain why b) does or doesn't exists. I believe it doesn't exist since ln(0) DNE, but I need explanation.
Let us take $$w_4=1-i = \sqrt{2}\left[\frac{1-i}{\sqrt 2}\right]$$ $$=\sqrt 2 e^{-i\frac{\pi}4} = e^{\color{red}{(\ln 2)/2 }}e^{i\color{green}{\left(-\frac{\pi}4\right)}}$$ $$=e^{\color{red} {a_4}}e^{i\color{green} {b_4}}$$
After EDIT:
Take for example $$w_3=-3i = 3e^{-\frac{\pi}2i}$$ $$=e^{\color{red}{\ln 3} }e^{\color{green}{-\frac{\pi}2} i}$$ $$=e^{\color{red}{a_3}}e^{i\color{green}{b_3}}$$