Solutions of an exponential function Find all solutions of $e^{z} = -1+i$
These are the things for what I did:
1) Let $z=x+iy$
2) $e^{x+iy} = e^{z} = √2(e^{i(3π/4+2kπ)})$
3) Equate moduli and arguments to see that: 
$x=ln(√2)$$y=\frac{3\pi}{4}+2k\pi$
and then after that I am lost, I have got no idea what to do next...
 A: If $z\in\mathbb C$, let $x\equiv\operatorname{Re}z$ and $y\equiv\operatorname{Im}z$, so that $z=x+iy$. Then, using Euler's formula, $$\exp(x+iy)=\exp(x)\exp(iy)=\exp(x)[\cos y+i\sin y]=-1+i.$$
Separating the real and imaginary parts, one must have
\begin{align*}
\exp(x)\cos(y)=&\,-1,\\
\exp(x)\sin(y)=&\,\phantom{-1}1.
\end{align*}
In particular,
\begin{align*}
\cos(y)=-\sin(y)=-\exp(-x).
\end{align*}
The first equality occurs if and only if $y=3\pi/4+2k\pi$ or $y=7\pi/4+2k\pi$ for some $k\in\mathbb Z$. But $\sin(y)=\exp(-x)$ must be positive, which leaves only $y=3\pi/4+2k\pi$, which implies also that $\sin(y)=1/\sqrt{2}$. Hence,
$$\sin(y)=\frac{1}{\sqrt{2}}=\exp(-x),$$
which implies that $x=(\log 2)/2$. Therefore, the set of solutions is
$$\left\{\left(\frac{\log 2}{2},\frac{3\pi}{4}+2k\pi\right)\,\Bigg|\,k\in\mathbb Z\right\}.$$
A: $$e^{x+iy}=-1+i$$
$$e^x\left(\cos(y)+i \sin(y)\right)=-1+i$$
$$\left\{ 
   \begin{array}{l l}
     e^x\cos(y)=-1 \\
     e^x\sin(y)=1
   \end{array} \right. $$
$$\left\{ 
   \begin{array}{l l}
     e^{2x}=2 \\
     \tan(y)=-1
   \end{array} \right. $$
with condition $\cos(y)<0$ and $\sin(y)>0$
$$\left\{ 
   \begin{array}{l l}
     x=\frac{1}{2}\ln(2) \\
     y=(2n+\frac{3}{4})\pi
   \end{array} \right. $$
$$z=\frac{1}{2}\ln(2)+i\:\left(2n+\frac{3}{4}\right)\pi$$
