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Question

Let $\mathbb{U}$ be the unit circle. Find all $P\in \mathbb{R}[X]$ such that $P(\mathbb{U})\subset \mathbb{U}$.

My attempt

My conjecture is that $P=\mu X^d$ for some $d\geqslant 0$ and $\mu \in \mathbb{U}$. I tried to prove it by contradiction, making a recurrence on the number of monomials in the polynom. Unfortunately it doesn't lead to anything interesting.

Could someone help me ?

NB : It isn't the same the question as Which polynomials fix the unit circle?, since $P(x+iy)\neq P(x)+iP(y)$ in general.

math
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  • I know it's been marked as a duplicate, but I'm not sure I understand what $P(\mathbb U)$ is supposed to mean when $P$ is a single-variable polynomial and $\mathbb U$ consists of complex numbers (or pairs of real numbers). – Sarvesh Ravichandran Iyer Mar 13 '22 at 09:40
  • @SarveshRavichandranIyer Actually it's not a duplicate. And $P(\mathbb{U})={P(z),z\in \mathbb{U}}$. – math Mar 13 '22 at 09:43
  • @Sit Exactly ! I hope one will reopen the post. – math Mar 13 '22 at 09:44
  • Are you allowed to use Complex Analysis to deal with this problem? – José Carlos Santos Mar 13 '22 at 09:49
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    @JoséCarlosSantos Actually this question doesn't come from any book or course. It's just a question I was asking myself so any solution is ok for me. – math Mar 13 '22 at 09:51
  • There might be another dupe though, at least by the answer here: https://math.stackexchange.com/questions/1233024 . So your conjecture seems to be correct but furthemore $\mu \in {-1,1}$ since $P\in \mathbb{R}[X]$. – Sil Mar 13 '22 at 09:57
  • @math Oh, I see , thanks for the clarification. I wasn't sure it was a duplicate but that's because I hadn't understood what $P(\mathbb U)$ was in the first place. – Sarvesh Ravichandran Iyer Mar 13 '22 at 11:58

1 Answers1

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Let $P(X)= \sum_{j=k}^n a_j X^j$, with $a_k \neq 0$ and $a_n \neq 0$. Then $$2\pi \overline{a_k}a_n=\int_{0}^{2\pi}P(e^{it})\overline{P(e^{it})}e^{i(n-k)}d=\int_{0}^{2\pi}|P(e^{it})|^2e^{i(n-k)}dt=\int_{0}^{2\pi}e^{i(n-k)}dt$$

since $|P(e^{it})|=1$ by hypothesis. Because $a_k \neq 0$ and $a_n \neq 0$, you get that necessarily, $n=k$ and $|a_n|=1$, which implies that $$\boxed{P(X)=\pm X^n}$$

TheSilverDoe
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