How to solve a Complex equation involving the conjugate: $(x+iy)^2-2(x-iy)+1=0$

Question

I want to find a Complex value for $z$ that satisfy the equation:

$$z^2-2z^*+1=0$$

But i have never seen the conjugate taking part of an equation.

What i have tried is give $z$ some components $x+iy$

So i have this: $(x+iy)^2-2(x-iy)+1=0$ And it reduces to this:

$$(x^2-2x-y^2+1)+i(2xy+2y)=0$$

But nothing seems to come out. It must be a simpler way, but i cant see it.

Answer 1

Hints:

Compare real parts and imaginary parts:

$$\begin{cases}x^2-2x-y^2+1=0\iff y^2-(x-1)^2=0\\{}\\2xy+2y=0\iff 2y(x+1)=0\end{cases}$$

Observe the second equation forces $\;y=0\;$ or $\;x=-1\;$ , so substitute in first equation both cases and etc.