Correct solution of a complex equation with modulus

Question

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.

Answer 1

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.

Answer 2

When $\text{z}\in\mathbb{C}$:

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

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:

$$ \begin{cases} \sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\left(3\sqrt{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}-2\right)=\Re^3\left[\text{z}\right]-3\Re\left[\text{z}\right]\Im^2\left[\text{z}\right]\\ 3\Re^2\left[\text{z}\right]\Im\left[\text{z}\right]-\Im^3\left[\text{z}\right]=0 \end{cases} $$