How to solve $z^n=\bar{z}$?

Question

$z^n=|z|^ne^{in\phi}=|z|e^{-i\phi}=\bar{z}$
$|z|^n=|z| \ \ \ \ \ \ \ \ \ $and $\ \ \ \ \ \ \ \ \ e^{in\phi}=e^{-i\phi}$
$|z|^{n-1}= 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n=\frac{-\phi}{\phi}=-1$

How to continue?

Answer 1

Equation $|z|^{n-1}$ can be used to determine the modulus of $z$, while $e^{in \phi}=e^{-i \phi}$ gives you $e^{i(n+1)\phi}=1$, from which one can deduce allowed values of $\phi$. Hint: there is more than one solution. You also wrote equation $n=- \frac{\phi}{\phi}=-1$, but this is wrong.