Find all solutions for $e^z = -2$ Find all solutions for $e^z = -2$. And we know there is no solution.

My trivial work:

Since if we take $\ln$ both sides, we simply have $z = \ln(-2)$ which isn't defined. 

But, since my work looks very trivial, I want to add more. 

Thank you.
By taking the natural log of both sides, one obtains $$ z = \log(2) + i\pi(2n+1)$$ $\forall n \in \mathbb{Z}$
Try to see if you can figure out why this is so.  

If it truly is for a complex analysis course the following will be helpful:
$e^z = e^{x+iy} = e^xe^{iy} = e^x(\cos(y) + i\sin(y))$
$e^z = -2$

$z = \ln(-2)$

Now since $e^{i\pi} = -1$,  $\ln(-1)=i\pi$,
and since $\log_z a + \log_z b = \log_z(ab)$,

$\ln(-2) = \ln(-1) + \ln(2)$

$\ln(-2) = i\pi + \ln(2)$

$z = i\pi +\ln(2)$