Complex numbers, finding solution for z How can you solve this? $z^2+2(1+i)z=2+2(\sqrt{3}-1)i$
I have tried to compare left and right side with real and imaginary part i then get
$ x^2+2x-y^2-2y=2$ 
$xy+x+y=(\sqrt{3}-1)$
But this equation can not be solved.
What else can i do? Setting in $z=re^{i\theta}$ does not help either
The answer is: $ \sqrt{3}-1$ and $-(1+\sqrt{3})-2i$

solving
$$
z^2+2(1+i)z-2-2(\sqrt{3}-1)i=0
$$
we have:
$$
z=-1-i\pm\sqrt{(1+i)^2+2+2(\sqrt{3}-1)i}=-1-i\pm\sqrt{2}\sqrt{1+\sqrt{3}i}
$$
now:
$1+\sqrt{3}i=2e^{i\pi/3}$ and
$$
\sqrt{1+\sqrt{3}i}=\sqrt{2}e^{i\pi/6}=\sqrt{2}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)
$$
so:
$$
z=-1-i\pm 2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i  \right)=-1-i\pm (\sqrt{3}+i)
$$
