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. 
 A: 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:
$e^z = e^{x+iy} = e^xe^{iy} = e^x(\cos(y) + i\sin(y))$
A: There is a solution.  I won't give it away, but I'll give you a hint:
$e^{i\pi} = -1$
So $e^z = -1$ has a solution.  Maybe you can jump from here to the solution for $e^z = -2$.
A: We have $$e^z = e^x\times e^{iy}.$$
So, we would like to have $e^xe^{iy} = 2\times(-1)$.
We can write $e^{iπ} = -1$ and thus $e^xe^{iy}= 2e^{iπ}$.  Comparing real and imaginary parts on both sides gives
$$e^x = 2 \implies x = \ln 2$$
and periodicity implies $y = \pi +2n\pi$, $n\in\mathbb{Z}$.
So, $$z = x+iy = \ln 2 + i(2n+1)\pi, \qquad n\in\mathbb{Z}.$$
A: $e^z = -2$
$z = \ln(-2)$
Now since $e^{i\pi} = -1$,  $\ln(-1)=i\pi$,
and since $\log_z a + \log_z b = \log_z(ab)$,
$\ln(-2) = \ln(-1) + \ln(2)$
$\ln(-2) = i\pi + \ln(2)$
$z = i\pi +\ln(2)$
A: Let e^z = 2 
z = ln2 + (θ+2kπ)
From e^z = e^(x+iy) 
e^x . e^iy = 2 
e^x(cosy+isiny)=2 
e^x cosy =2 
which means θ=y=0 
so z= ln2 +2kπ

and because e^iπ=-1, 
e^z(e^iπ)=2(-1)=-2 
e^(iπ+ln2+2kπ)= -2
