Q:

Solve a complex equation with conjugate and square

This seems like a rather simple equation:
$$z^2 = \bar{z}$$
I tried $z=x+iy$ and then
$$x^2-y^2 + 2xyi=x-iy,$$
but this is an equation in two variables. Does anyone have a clever hint without giving the entire solution away? I want to solve it myself.

A:

Note that$$z^2=\overline z\implies|z|^2=\bigl|\overline z\bigr|=|z.|$$Therefore, $|z|=0(\iff z=0)$  or $|z|=1$. It turns out that $0$ is indeed a solution. If $|z|=1$, then $\overline z=\frac1z$. So, solve the equation $z^2=\frac1z(\iff z^3=1)$.

