How do I solve $e^{iz} = 3i$? I am stuck on the question that asks me to solve $e^{iz} = 3i$. 
I rewrote $e^{iz}$ as $e^{ix-y}$ and thus as $e^{ix}\cdot e^{-y}$ and thus as $e^{-y}\cdot (\cos x + i\sin x)$. 
I then wrote $e^{-y}\cos x = 0$ and $e^{-y}\sin x = 3$. When I try to solve this, however, I find $x = \pi/2 + 2k\pi$ as solution for the first expression, but I can't find a solution for $e^{-y}\sin x = 3$, because there are no solutions for $\sin x = 3$.
I then tried to find y through solving $e^{-y}\cdot \cos(\pi/2 + 2k\pi)=0$, this however gives me $y = \log 0$, which has no solutions. What am I doing wrong? Thanks!
You're doing good: if $z=x+iy$, then $e^{iz}=e^{-y}(\cos x+i\sin x)$, so you have
$$
\begin{cases}
e^{-y}\cos x=0 \\[4px]
e^{-y}\sin x=3
\end{cases}
$$
From the first equation you get $\cos x=0$, so $x=\pi/2+2k\pi$ or $x=-\pi/2+2k\pi$. Substituting in the second equation, the former gives
$$
e^{-y}=3
$$
so $y=-\log 3$. The latter instead gives $-e^{-y}=3$, which has no solution.
Use $z=x+iy$ to obtain

$$\exp(iz)=\exp(ix-y)=\exp(-y)\exp(ix)=\exp(-y)\cos(x)+i\exp(-y)\sin(x)=3i$$

By comparing the real and imaginary part we obtain two simultaneous equations:

$$\exp(-y)\cos x = 0 \implies \cos x = 0 \text{ as }\exp(-y)\neq 0$$
$$\exp(-y)\sin x = 3.$$