I am trying to find the algebraic fractions $F$ of $\mathbb{C}(X)$ with no roots on $\mathbb{U}$ that stabilizes $\mathbb{U}$:
$$ \forall u \in \mathbb{U},\quad F(u) \textrm{ exists and } F(u) \in \mathbb{U} $$
I have already solved the case of $F \in \mathbb{C}[X]$: it is the $aX^n$ with $a \in \mathbb{U}$ and $n \in \mathbb{N}$.
It is clear that the $aX^k$ with $a \in \mathbb{U}$ and $k \in \mathbb{Z}$ are solutions. How to show they are the only solutions?
I tried to mimic the proof of the polynomial problem: let $F \in \mathbb{C}(X)$ be a solution. Let us write $F = \tfrac{P}{Q}$ with $P, Q$ polynomials of $\mathbb{C}[X]$ such that $Q \neq 0$, $Q(z) \neq 0$ on $\mathbb{U}$ and $\textrm{gcd}(P, Q) = 1$. We easily see $P \neq 0$. We have:
$$ \forall u \in \mathbb{U},\quad F(u) \overline{F(u)} = \frac{P(u) \overline{P(u)}}{Q(u) \overline{Q(u)}} = 1 $$
As $\overline{P(u)} = \overline{P}(\overline{u}) = \overline{P}(\frac{1}{u})$, denoting $N := \max(\deg P, \deg Q)$, we have $P(u) R(u) = Q(u) S(u)$ on $\mathbb{U}$ for the polynomials $R(X) = \overline{P}(1/X)X^N$ and $S(X) = \overline{Q}(1/X)X^N$. As $\mathbb{U}$ is infinite, the polynomial equality holds: $PR = QS$. As $P$ is prime with $Q$, $P$ divides $S$. Same holds for $Q$ and $R$. But I do not know how to go further.