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 $-(1+\sqrt{3})-2i$
 A: By setting $z^2 + 2(1+i)z-2-2(\sqrt{3}-1)i = 0$, you can see this is a second degree polynomial, we can solve this by using the standard formula for the solution of second degree polynomials (which is also valid in $\mathbb{C}$). You will have to find two roots of the discriminant, and this one can indeed do by using polar coordinates(or maybe things will turn out more nicely and you can see the square root inmediately). 
A: I didn't see this anywhere above, except there was one polar substitution above, but only briefly.  So here goes.
We can rewrite in polar form as
$$z^2+2\sqrt{2}e^{i\pi/4}z=4e^{i\pi/3}-2e^{i\pi/2}$$
Completing the square by adding $2e^{i\pi/2}$ to both sides (which will cancel the last term), we get
$$\left(z+\sqrt{2}e^{i\pi/4}\right)^2=4e^{i\pi/3}$$
$$z+\sqrt{2}e^{i\pi/4}=\pm2e^{i\pi/6}$$
$$z=-\sqrt{2}e^{i\pi/4}\pm2e^{i\pi/6}$$
at this point, the conversion to rectangular neatens it up
$$z=(-1+i)\pm(\sqrt{3}+i)$$
and thus
$$z=(\sqrt{3}-1)+2i \qquad \qquad z=-1-\sqrt{3}$$
A: It's a quadratic, use the quadratic formula
$$z = \frac{ -2(1+i) \pm \sqrt{(2(1+i))^2+4(2+2(\sqrt{3}-1)i)}}{2}$$
Now you have to simplify.
A: What about the high school formula for the roots of a quadratic? We have
$$z^2+\left[2(1+i)\right]z-2\left(1+(\sqrt3-1)\right)i=0\;\implies\;$$
$$\Delta =(2(1+i))^2+4\left[2\left(1+(\sqrt3-1)\right)\right]=4\left(2i+2+2\sqrt3-2\right)=8\left(\sqrt3+i\right)$$
Continue on and observe that above you almost has the square root of $\;i\;$ ...
A: 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:
$$
z=-1-i\pm 2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i  \right)=-1-i\pm (\sqrt{3}+i)
$$
