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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.

John
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  • Small point : we must assume $r > 0$, else $D = {0},{1}$ also are allowed. – Sarvesh Ravichandran Iyer Mar 22 '20 at 10:35
  • We could eliminate ${0,1}$ (and some other discrete sets) by noting that these lack the rotational symmetry of a disc. – Oscar Lanzi Mar 22 '20 at 11:16
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    It's easy to see that $D$ is contained in the unit disc. Since $D^n = D$ for all $n$, we must have $|z^n| = |z|^n$ be bounded for all $z \in D$. This implies $|z| \le 1$. – user759562 Mar 22 '20 at 11:26
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    As @user759562 said it is easy to prove that $z_0 < 1$. I can also prove that there is at least one point $w$ on the disc such that $|w| = 1$. If $w \neq 1$ then $w^2$ is also on the disc, and because we have a circle that is completely contained in another circle and shares 2 boundary points, it is not difficult to see that they are the same circle. The only thing I can't seem to prove is the case when $w = 1$ – Boxonix Mar 22 '20 at 14:09
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    @Boxonix That's where I got to too. Note also that $0$ must lie in $D$, because there must be some $z \in D$ with $|z| < 1$, and so $z^n \in D$ converges to $0$. Since $D$ is closed, $0 \in D$. Thus, we can say that $B[1/2;1/2] \subseteq D \subseteq B[0; 1]$. – user759562 Mar 22 '20 at 15:16

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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 $|-r+2irz_O|>r$.

Berci
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