Work in polar form.

$\displaystyle \overline{z}=z^2$,

so:

$\displaystyle |\overline{z}|=|z|=|z^2|=|z|^2$

so for non-zeros solutions $\displaystyle |z|=1$, and $\displaystyle z=e^{i\theta}$ for some $\displaystyle \theta \in [0,\2\pi)$

Then:

$\displaystyle \overline{z}=e^{-i\theta}=z^2=e^{2i\theta} $

etc

CB