When does a cubic equation has two roots with same absolute values Let $$z^3+bz^2+cz+d=0$$ be a cubic equation with complex coefficients. Suppose $z_1, z_2$ and $z_3$ are its roots.
I need to find a condition on $b,c,d$ so that $$|z_1|=|z_2|.$$ How can I find such a relationship? Any help would be appreciate. Thank you.

If $|z_1| = |z_2|$, you can write $z_1 = u e^{i\theta}$ and $z_2 = u e^{-i\theta}$ for some real $\theta$ and complex $u$.  The cubic is then
$$(z - u e^{i\theta})(z - u e^{-i\theta})(z - z_3)= (z^2 - 2 u \cos(\theta) + u^2)(z - z_3)$$
We then get
$$ \eqalign{b &= -2 u \cos(\theta) - z_3\cr
            c &= 2 u z_3 \cos(\theta) +u^2\cr
            d &= -u^2 z_3\cr} $$
EDIT:
 We can eliminate the complex parameters $u$ and $z_3$ to obtain
an equation involving $b,c,d$ and $\cos(\theta)$:
$$ 64\,{d}^{2}  \cos^6 \left( \theta \right)  -16\,d
 \left( bc+3\,d \right)  \cos^4 \left( \theta \right)  + 4\, \left( {b}^{3}d-bcd+{c}^{3}+3\,{d}^{2} \right)   \cos^2
 \left( \theta \right)  -(bc - d)^2 = 0
$$
If we write $\cos^2(\theta) = v$, that is a cubic in $v$, and a condition (maybe the only one needed) on $b,c,d$ is that this cubic has a solution in the real interval $[0,1]$.
