# Solve this system of equations with modulus in $\mathbb{C}$

I tried to solve this exercise but failed after some calculations, tough I've extracted some important piece of information from it. The system is:

$$\left\{\begin{array}{c} x|y| = z^2 \\ y|z| = x^2 \\ z|x| = y^2 \end{array}\right.$$ where $$x,y,z \in \mathbb C$$. First I tried substituting $$z^2$$ with $$\left(\frac{y^2}{|x|}\right)^2 = \frac{y^4}{x\cdot \overline{x}}$$ but I soon gave this idea up as it caused much harder equations to appear. But the I thought of multiplying them all together and I found the following: $$xyz|x||y||z| = x^2y^2z^2 \\ xyz|xyz| = (xyz)^2 \\ |xyz| = xyz \\ Im(xyz) = 0 \leftrightarrow xyz \in \mathbb R$$

After this I am stuck. What can I do next to solve the system?

First note that if $$x, y, z$$ are solutions to your system, then you must have $$|x| = |y| = |z|$$; you can see this by taking the absolute value of each of your equations, and considering the equation where the largest of the three has its square on the right hand side.
In particular -- setting aside the solution $$x = y = z = 0$$ -- note that for any solution $$x, y, z$$, and any positive $$\alpha$$, also $$\alpha x, \alpha y, \alpha z$$ is a solution. Thus we can take $$\alpha = 1/|x|$$ and reduce to the case where $$|x| = |y| = |z|$$, which considerably simplifies your equations to $$x = z^2,\\ y = x^2,\\ z = y^2.$$ Now we can do repeated substitution, and obtain that $$x^7 = y^7 = z^7 = 1$$. Each of $$x, y, z$$ is a 7th root of unity. And now you can simply check: if $$x$$ is any seventh root of unity, and we set $$y = x^2$$, and $$z = y^2$$, then the latter two equations are clearly satisfied, and $$z^2 = y^4 = x^8 = x$$, so therefore so is the first.
Hence the solutions to the equation are $$\alpha \times \exp(2\pi i \tfrac k7)$$, for $$k \in \{0,1,2,3,4,5,6\}$$, and $$\alpha$$ non-negative real.
• $$x|y| = z^2 \implies |x| |y| = |z|^2$$, which implies $$|xyz|=|x|^3=|y|^3=|z|^3$$.
• $$x|y| = z^2 , y|z| = x^2 \implies x^3 = y z^2$$