Let $D= \{z \mid\vert z-z_O\vert \leq r \}$ be a closed disc in the complex plane and $D^2=\{z_1z_2 \mid z_1,z_2 \in D \}$ . Prove that if $D=D^2$, $D$ is a unit disc and $z_O=0$.
I really have no idea how to start this problem, $z_1z_2$ might suggest something related to the trigonometric form but it doesn't seem helpful. Maybe there is a solution based on some sort of geometrical interpretation, or something that is related to the sets $D$ and $D^2$, but I honestly don't know what to do.
First, $D\subseteq \{z:|z|\le 1\}$, because it can't contain any number $z$ with $|z|>1$, since then it would contain $z^n$, and thus $D$ would be unbounded, which it isn't.
Let $z\in D$ be an element with maximal absolute value (such exists since $D$ is compact). If $|z|<1$, then each element of $D^2$ has absolute value $\le |z|^2 <|z|$, so in particular, $z\notin D^2$, contradicting to $D^2=D$.
So, we get $|z|=1$. If we prove $z_O=0$, we are done.
So, assume $z_O\ne 0$.
Since $D$ is a disk, its points $w$ satisfy $|w|\le\, |w-z_O|+|z_O|\le r+|z_O|=|z|$ where the $z\in D$ with biggest absolute value is unique and it is explicitly $$z=\left(\frac r{|z_O|}+1\right)z_O\,. $$ As was noted in the comments, we can easily exclude $z\ne 1$, as then $z \ne z^2 \in D$ contradicting the uniqueness of $z$ with maximal absolute value.
Finally, $z=1$ means $z_O>0$, and then $z_O+r=1$, and $(z_O+ir)^2 =z_O^2-r^2+2irz_O=(z_O-r)+2irz_O\notin D$ because $|-r+2irz_O|>r$.