find all value of z belong to C Find all values of $z\in\mathbb{C}$ such that 
  
  $$ e^z = -3i$$


My try:

$$e^z=e^{x+iy}$$
$$-3i = re^{i\cdot0}$$

take $\ln$ both sides 

$$z =\ln(-3i)$$

stuck here, need help to solve it.
Note that
$$
-3i=3(-i)=3e^{i\frac32\pi+i2k\pi},\;\; k\in\Bbb Z
$$
and
$$
e^z=e^xe^{iy}.
$$
Thus
\begin{align*}
e^z=-3i\;\;
&amp;\Longleftrightarrow\;\;e^x=3\;\;\vee\;\; e^{iy}=e^{i\frac32\pi+i2k\pi}\\
&amp;\Longleftrightarrow\;\;x=\log3\;\;\vee\;\;y=\frac32\pi+2k\pi\\
&amp;\Longleftrightarrow\;\;z=\log3+i\left(\frac32\pi+2k\pi\right),\;\;k\in\Bbb Z.
\end{align*}

Finally be careful using logarithims when you're dealing with complex numbers: here the log function is different than the one we use with real numbers. Take a look at the Complex Logarithm.
$z=a+ib$$e^z = e^a*e^{ib}$
$e^a*e^{ib}=-3i$
Magnitude of $e^{ib} = 1$
So, $e^a=3$, $a = ln3$
$e^{ib}=-i$ 
$b=$$(2n+1)\pi\over 2$ where $n$ is an integer.