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!
 A: 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.
A: Your solution is fine, until you get to the part about $e^{-y}\sin x = 3$. Since you already know that $x=\pi/2 + 2k\pi$, then you know that $\sin x=1$, so you just need to solve $e^{-y}=3$.

If I were doing this problem, I'd probably skip all the $x,y$ stuff, and start out by writing $3i$ in polar form, and then rewriting the modulus as an exponential: $$3i=3e^{i(\pi/2+2k\pi)}=e^{\ln 3}e^{i(\pi/2+2k\pi)}=e^{\ln 3+\pi i(\pi/2+2k\pi)}.$$ Now you're trying to solve: $$e^{iz}=e^{\ln 3+i(\pi/2+2k\pi)}.$$ Equating exponents, you get a straightforward equation. If you only need one solution, you can leave off the $2\pi i$ part, i.e., set $k=0$.

Even more briefly, if you just think of the function $e^z$ as using the real part of $z$ to determine a modulus, and the imaginary part to determine an argument, then you think, "the real part of $iz$ must be $\ln 3$, and the imaginary part must be equivalent on the circle to $\pi/2$". Then you can just write down: $$iz=\ln 3 + i(\pi/2+2k\pi),$$ and then divide by $i$.
A: Hint $$|e^{iz}|=e^{-y}=|3i|=3$$
A: Others have discussed your approach. To avoid the issues altogether
you could take logs (base e) where $\log{z}=\log{|z|}+i(\arg{z}+2k\pi)$ thus
$$
\log(e^{iz})=\log(3i)\\
\Rightarrow iz=\log3+i(\frac{\pi}{2}+2k\pi)\\
\Rightarrow z=\frac{\pi}{2}+2k\pi-i\log3
$$
A: 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:
$$\exp(-y)\cos x = 0 \implies \cos x = 0 \text{ as }\exp(-y)\neq 0$$
$$\exp(-y)\sin x = 3.$$
