Find all the solutions to the equation: $e^z = e^{-1 + i\pi}$ Im slightly stuck on hiw to attampt this...
I thought to write both sides in polar form, 
ie.
$$e^x (\cos(y) + i\sin(y)) = e^{-1}(\cos(\pi)+i\sin(\pi))$$
I could equate it from here i think, but it specifically says in the question find ALL the solutions to the equation.
By doing it this way, i think i would only find one.
I know i need to add $2k\pi$ in at some point. 
Any help would be much appreciated.

The exponential map is 1-1 on the strip $\{x+iy : -\pi \le y < \pi \}$ and $e^z = e^{z+2n\pi i}$ for all $n \in \mathbb{Z}$. Since $z = -1 +i\pi$ is one solution of it. So, $z = -1 + i(\pi +2n\pi)$ where $n \in \mathbb{Z}$ are all the solutions of your equation.
