Solving a complex number equation with both $z$ and its conjugate $\bar z$

Question

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!

Answer 1

Denote $z=a+bi$ with $a,b\in\mathbb{R}$, then your equation says: $$4(a+bi)=(a-bi)^2.$$ Which means that $$4a+4bi=(a^2-b^2)-2abi.$$ Now the above equality of complex numbers is a system of equations of real numbers: $$\begin{cases} 4a=a^2-b^2\\ 4b=-2ab \end{cases}.$$ Solve it and you'll find the solutions.

Answer 2

Hint: use the polar representation $z = r e^{i\theta}$.

Answer 3

There is the obvious solution $z=0$. From now on, assume $z\ne 0$. Taking norms, we find that $|z|=4$. Let $z=4e^{i\theta}$. Then we want $e^{i\theta}=e^{-2i\theta}$. We leave the rest to you.