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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.

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    $\begingroup$ You could try reverse engineering: start with a factored cubic, and assuming two roots have the same modulus, see what happens when you multiply (note that complex conjugates have the same modulus, which certainly gives a large class of the cubics in question). $\endgroup$ – pjs36 Jan 19 '16 at 6:49
  • $\begingroup$ Any thoughts on the answers that have been posted? $\endgroup$ – Gerry Myerson Jan 21 '16 at 11:41
  • $\begingroup$ Earth to Nilan, come in please. $\endgroup$ – Gerry Myerson Jan 23 '16 at 4:25
  • $\begingroup$ @GerryMyerson: :) $\endgroup$ – Bumblebee Jan 28 '16 at 3:41
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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 $[0,1]$.

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    $\begingroup$ Shouldn't it be $2u\cos(\theta)z$? $\endgroup$ – amcerbu Jan 19 '16 at 7:31
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    $\begingroup$ You have only considered the case when $z_1$ is the conjugate of $z_2$. What about the case of repeated roots, i.e., $z_1 = z_2$ $\endgroup$ – user297008 Jan 19 '16 at 7:50
  • $\begingroup$ @Tanuj, that's $\theta=0$. $\endgroup$ – Gerry Myerson Jan 19 '16 at 8:37
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    $\begingroup$ @GerryMyerson The problem is that you have defined it such that a) $b,c,d\in\mathbb{R}$, and b) $z_1=-z_2$, which is a direct result of your assumption that $z_1$ and $z_2$ have opposite $\theta$'s. $\endgroup$ – Simply Beautiful Art Jan 19 '16 at 13:16
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    $\begingroup$ @Simple, I can't speak for OP, but we can take $u$ to be either one of the solutions to $u^2=z_1z_2$, and then get $\theta$ from $e^{i\theta}=z_1/u$. Or, we can first get $\theta$ from $z_1/z_2=e^{2i\theta}$, and then get $u$ from $u=z_2e^{i\theta}$. $\endgroup$ – Gerry Myerson Jan 19 '16 at 22:05
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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:

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

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

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