When does a cubic equation has two roots with same absolute values

Question

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.

Answer 1

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 $[0,1]$.

Answer 2

In comparison to Robert Israel's answer, I note the following:

$$|ue^{\alpha\cdot i}|=u$$

for $\alpha\in\mathbb{R}$ and $u>0$.

Generally, Euler form of a complex number has the magnitude of the complex number as a coefficient and the real angle multiplied by $i$ with base $e$, which allows us to encompass all possible complex numbers with a slight exception of maybe $0+0i$.

Now, we can define our problem such that we have:

$$z_1=ue^{xi}, z_2=ue^{yi}$$

So now we have:

$$(z-ue^{xi})(z-ue^{yi})(z-z_3)$$

And you could multiply them all out if you want.

Another answer goes along the line of finding the root of a cubic polynomial, which may be found on Wolfram|Alpha.

Take the absolute value of the thing and try to solve using definitions of the absolute value operation:

$$|z|=|a+bi|=\sqrt{a^2+b^2}$$

Or:

$$|z|=z, z\ge0$$

$$|z|=-z,z\le0$$