solutions of $az^2 + bz + c = 0$ have the same module $r > 0$

Question

Prove that the solutions of the equation $az^2 + bz + c = 0$, with $a, b, c \in \mathbb{C}$ have the same module $r > 0$ if and only if $b\overline{c} = r ^ 2 a\overline {b}$.

How can I deal with it? I have tried a rather "brute" solution, but it did not work and I am out of ideas...

$r^2 = \frac{b^2\overline{ac}}{ab\overline{ab}}$ and $r = |-b - \sqrt{b^2 -4ac}| = |-b+\sqrt{b^2-4ac}$ and I tried proving it, but it gets quite complicated.

Answer 1

let $p,q$ be the roots of equation. then $$\bar p=\frac{r^2}{p},\bar q=\frac{r^2}{q}$$ Now by vieta $$\color{blue}{p+q=\frac{-b}{a}},pq=\frac{c}{a}$$ hence taking conjugate of blue colored equation $$\color{blue}{\frac{-\overline{ b}}{\bar a}=\bar p+\bar q}=r^2(\frac{p+q}{pq})=\frac{-r^2b}{c}$$ $$\iff b\bar c=r^2a \bar b$$