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.

Using Robert's hint, note that your equation is equivalent to
\begin{align*}
e^{z + 1 - i\pi} = 1.
\end{align*}
Hence $z+1-i\pi = 2\pi i n$ for some integer $n$. Solving for $z$, we find that
\begin{align*}
z = 2\pi i n + i \pi - 1 = i\pi (2n + 1) - 1,
\end{align*}
for integers $n$.
