this is the process I follow:
then i get: $Im(z)=2y/\sqrt{x^2 + y^2}$ and $Re(z)=(-(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
what should i do next?
You may simplify the notation from the beginning. Call $z=a+bi$ and your equation becomes:
$$a^2+b^2+(a+bi)\sqrt{a^2+b^2}+2=2bi$$ or
$$[a^2+b^2+a\sqrt{a^2+b^2}+2]+[bi\sqrt{a^2+b^2}]=2bi$$
So, you can split it into the real part and the imaginary part.
$$a^2+b^2+a\sqrt{a^2+b^2}+2=0\quad (1)$$
$$b\sqrt{a^2+b^2}=2b \quad (2).$$
From $(2)$ you get $b=0$ or $\sqrt{a^2+b^2}=2$.
Can you finish?
Write $z=re^{i\theta}=r(\cos\theta + i \sin\theta)$ with $r,\theta\in\Bbb{R}$; your equation becomes $r^2+r\times r(\cos\theta+i\sin\theta)+2=2ir\sin\theta$. Identifying real and imaginary parts of the left & right members, you get: $\begin{cases} r^2(1+\cos\theta)+2=0&(1)\\[5pt] r^2\sin\theta=2r\sin\theta&(2) \end{cases}$
Now (1) implies that there is no solution to the problem… (Or did I make a mistake?)
We can start by looking at the real and imaginary parts separately, inspired by the fact that the LHS has a lot of things that are purely real while the RHS is purely imaginary. So focusing on the imaginary part gives us:
$$\begin{eqnarray}Im(|z|^2 + z|z| + 2) & = & Im(2i Im(z)) \\ Im(z)|z| & = & 2 Im(z) \\ \implies |z| = 2 & \lor & Im(z) = 0\end{eqnarray}$$
If we check the $Im(z) = 0$ case first, that means $z \in \mathbb{R}$, and the equation becomes:
$2z^2 + 2 = 0 \implies z^2 = -1$
which has no solutions over the reals.
So the only possibly solutions are where $|z| = 2$, and if we put that into the equation we get:
$$\begin{eqnarray}2^2 + 2z + 2 & = & 2i Im(z) \\ z + 3 & = & 2i Im(z) \\ Re(z + 3) & = & Re(2i Im(z)) \\ Re(z) + 3 & = & 0 \implies Re(z) = -3 \end{eqnarray}$$
but this is a problem, because $|z|^2 \geq Re(z)^2$ which would mean $2^2 \geq (-3)^2$ which is clearly a contradiction.
So in fact there are no solutions to this equation.
