Find all $z\in\mathbb{C}$ satisfying $z^2 = |z|^2$ Let $z=x+yi$, I get $y=xi$ finally, but what's the next step?
 A: Note that $|z|^2 = z\bar z$.  So, we have
$$
z^2 = |z|^2\\
z^2 - z\bar z = 0\\
z(z - \bar z) = 0
$$
so: we must have $z = 0$, or $z = \bar z$ (which is to say that $Im(z) = 0$).
A: Clearly $|z|^2$ is real and non-negative. So you need $z^2$ real and non-negative. That happens if $z$ is real, and in that case it's clear that $z^2=|z|^2$. Finally, observe that if $z$ is not real, then $z^2$ either is not real or is negative.
So the solution is: $z$ is real.
A: There isn't really any need to break down $z$ into real parts or polar form or multiple cases. Firstly $|z|^2=z\bar z$, so you are looking at the equation $z\bar z = zz$.
Either $z=0$ or else you can cancel $z$ from both sides, whence $z=\bar z$, and in either case that means $z$ is real.
So, only real numbers will work, and it's clear they all work.
A: Of course $z=0$ is a solution. If $z\ne0$, write $z=|z|u$, where $|u|=1$. Then the equation becomes
$$
|z|^2u^2=|z|^2
$$
Can you say what $u$ should be and finish?
A: Since $z^2=|z|^2 e^{2i\theta}$, for some $\theta$, then the equation is $e^{2i\theta}=1$. Hence $\theta \in \pi\mathbf{Z}$, i.e., $z \in \mathbf{R}$.
A: In polar coordinates you get $(r(\cos \theta + \sin \theta))^2 = r^2(\cos 2\theta + \sin 2\theta) = r^2 = |r(\cos \theta + \sin \theta)|^2$ so $\theta \in \pi \Bbb Z$ and therefore $z \in \Bbb R$
