Having $x^3=(3+4i)\overline{x},$ with $x$ a complex number, what is the sum of the moduli of its roots?

Question

I've started by applying modulus on both sides knowing that $$|x| = \left\lvert\overline{x}\right\rvert=\sqrt{a^2+b^2},$$ which is a constant $k.$ That leads me to the equation $k^3=5k,$ which leads to either $k=-\sqrt{5}$ or $k=\sqrt{5}.$ The sum of the moduli of the roots will be just $2\sqrt{5}$?

Answer 1

No, it is not $2\sqrt5$, and your justification is not correct. The number $k$ cannot be $-\sqrt5$, since $k\geqslant0$.

There is an obvious root, which is $0$. If $x$ is a root and $x\neq0$, then\begin{align}x^3=(3+4i)\overline x&\implies\lvert x\rvert^3=5\left\lvert\overline x\right\rvert=5\lvert x\rvert\\&\implies\lvert x\rvert^2=5.\end{align}Therefore, $x=\sqrt5\bigl(\cos(\theta)+i\sin(\theta)\bigr)$ for some $\theta\in\mathbb R$. And\begin{align}x^3=(3+4i)\overline x&\iff5\sqrt5\bigl(\cos(3\theta)+i\sin(3\theta)\bigr)=\sqrt5(3+4i)\bigl(\cos(-\theta)+i\sin(-\theta)\bigr)\\&\iff\cos(3\theta)+i\sin(3\theta)=\left(\frac35+\frac45i\right)\bigl(\cos(-\theta)+i\sin(-\theta)\bigr).\end{align}Now, if $\alpha\in\mathbb R$ is such that $\cos\alpha=\frac35$ and that $\sin(\alpha)=\frac45$, the last equality is equivalent to$$\cos(3\theta)+i\sin(3\theta)=\cos(\alpha-\theta)+i\sin(\alpha-\theta).$$ This leads to $4$ solutions: $\theta=\frac\alpha4$, $\theta=\frac\alpha4+\frac\pi2$, $\theta=\frac\alpha4+\pi$, and $\theta=\frac\alpha4+\frac{3\pi}2$ (after this, you will be getting the same roots). So, besides $0$, there are $4$ roots, all of which have absolute value $\sqrt5$. Therefore, the answer is $4\sqrt5$.

Answer 2

You're correct that if $k=|x|,$ then $x^3=(3+4i)\overline{x}$ yields $k^3=5k$ by taking the modulus of both sides. However, from there, you've gone off track.

From $k^3=5k,$ we can conclude that $k\in\left\{0,-\sqrt{5},\sqrt{5}\right\}.$ However, $k$ is a modulus, so cannot be negative! Thus, $k\in\left\{0,\sqrt{5}\right\}.$

If the equation were a polynomial in $x,$ we could conclude that it has exactly three roots, and at that point, you'd be able to conclude that the sum of the roots' moduli is $2\sqrt{5},$ so long as you could show that $0$ was a root of multiplicity $1.$ However, it is not a polynomial in $x.$

By the prior work, we know that $|x|=0$ or $x=\sqrt{5},$ so, recalling that $x\overline x=|x|^2,$ we find that $$x^3=(3+4i)\overline{x}$$ holds if and only if either $x=0$ or $$x^4=(3+4i)x\overline{x}=(3+4i)|x|^2=15+20i.$$ Thus, since $x^4=15+20i$ is a polynomial in $x$ whose roots necessarily have modulus $\sqrt{5},$ the sum of the moduli is $0+4\sqrt{5}=4\sqrt{5}.$

Answer 3

Let $x=re^{it}, \bar x=re^{-it}$ where $r\ge0, t$ are real

$r^3e^{3it}=(3+4i)re^{-it}$

If $r\ne0,$ $$r^2e^{i4t}=3+4i$$

Taking modulus both sides $r^2=5\implies r=\sqrt5$

$\implies e^{i4t}=\dfrac{3+4i}5=e^{iy}$ where $y=\arccos\dfrac35$

$\implies 4t=2m\pi+\arccos\dfrac35$ where $m$ is any integer

$\implies t=\dfrac{m\pi}2+\dfrac{\arccos\dfrac35}4$ where $m=0,1,2,3$

So, we have $4$ unequal non-zero roots each having modulus $=\sqrt5$