Determine all possible values of $z\in\mathbb{C}$ that satisfy the equation $4z = \overline{z}^2$.
Where $\overline{z}$ represents the complex conjugate.
(Hint: There are $4$ solutions.)
Observations
If we had $4z=z^2$, that would be an easy quadratic equation, with solutions $0,4$.
And if it was $4\bar z = \bar z^2$, then after substitution $\zeta=\bar z$ we have a quadratic equation again.
But this equation has both $z$ and $\bar z$. I'm not sure how to solve these types of problems. Any tips or how to do these would be great thanks!
Best Answer
Hint: use the polar representation $z = r e^{i\theta}$.