Correct solution of a complex equation with modulus I struggle to solve a complex equation with modulus.
This is the text:
$$  \vert z\vert (3\vert z\vert -2 ) = z^3$$
One way to resolve this is substitute $\vert z\vert$ with $\sqrt {a^2+b^2}$, solve the cube etc, and after that set, in a system, the real part = 0 and the imaginary part = 0. This method should 100% work, but another way must exist.
I read about a second way: I consider the 2nd part of the equation, and rewriting it in the trigonometric way, I can say that cis(3theta) must be a real number, because the first term of the equation is a real number too. So, cis(3theta) must be = +-1. Is this lecit?
Thanks in advance, I perfectly know that the first method is enough but the second coul be more aesthetic.
 A: You are close.
In polar coordinates
$$r(3r-2)=r^3e^{i3\theta}$$ or
$$3r-2=r^2e^{i3\theta}$$ after noting that $z=0$ is a solution.
Then $e^{i3\theta}$ is indeed real positive or negative, giving
$$+\to r^2-3r+2=0\implies r=1,2\times 1,\omega,\omega^2$$
$$-\to r^2+3r-2=0\implies r=\frac{\sqrt{17}-3}2\times -1,-\omega,-\omega^2.$$
Thus there are ten solutions.
A: When $\text{z}\in\mathbb{C}$:


*

*$$\left|\text{z}\right|=\sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}$$

*$$\text{z}^3=\Re^3\left[\text{z}\right]-3\Re\left[\text{z}\right]\Im^2\left[\text{z}\right]+\left(3\Re^2\left[\text{z}\right]\Im\left[\text{z}\right]-\Im^3\left[\text{z}\right]\right)i$$


So,
$$\left|\text{z}\right|\left(3\left|\text{z}\right|-2\right)=\text{z}^3\Longleftrightarrow$$
$$\sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\left(3\sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}-2\right)=\Re^3\left[\text{z}\right]-3\Re\left[\text{z}\right]\Im^2\left[\text{z}\right]+\left(3\Re^2\left[\text{z}\right]\Im\left[\text{z}\right]-\Im^3\left[\text{z}\right]\right)i$$
We can set up a system:
$$
\begin{cases}
\sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\left(3\sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}-2\right)=\Re^3\left[\text{z}\right]-3\Re\left[\text{z}\right]\Im^2\left[\text{z}\right]\\
3\Re^2\left[\text{z}\right]\Im\left[\text{z}\right]-\Im^3\left[\text{z}\right]=0
\end{cases}
$$
