A simple cubic equation problem:

Question

Consider the cubic equation $$az^3-bz^2+\bar{b}z-\bar{a}=0$$ where $a$ and $b$ are non-zero complex numbers. Suppose $z_1, z_2$ and $z_3$ are the roots.

Question: Which $a$ and $b$ gives $|z_1|=|z_2|=|z_3|=1$ ?

We can easily prove $|z_1z_2z_3|=1$ and if $z$ is a root, then $\dfrac{1}{\bar{z}}$ is also a root. This implies that there is a root with absolute value $1.$
But after that how can we continue ?

Answer 1

For $z^{3}+w z^{2}+\bar{w} z+1=0$, there's a paper discussed. Still need to check for your case.