$e^z=-3i$ find $z\in \mathbb C$ check my answer

Question

I am unsure of my solution to this question, since the definition of the complex logarithm is somewhat complex.

Since $-3i = 3e^{i\frac{3}{2}\pi}$ we get that $e^z=3e^{i\frac{3}{2}\pi}$

So if we use logarithm function like we did in $\mathbb R$ we would get

$z=\ln(3e^{i\frac{3}{2}\pi})=\ln3+\ln(e^{i\frac{3}{2}\pi})=\ln3+i\frac{3}{2}\pi$

Is this the correct answer? Are there more correct answers?

Answer 1

An idea: putting $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$

$$-3i=e^{z}=e^x\cos y+ie^x\sin y$$

Now compare real and imaginary parts and stuff.