We denote by $\mathcal{R}_{n}$ the set of rational functions with no pole in $\mathbb{U}$ of the form $\frac{P}{Q}$ where $P$ and $Q$ are two elements of $\mathbf{C}_{n}[X]$.
Let $F \in \mathcal{R}_{n}$, $P$ and $Q$ be two elements of $\mathbf{C}_{n}[X]$ satisfying $F = \frac{P}{Q}$ and $$\forall z \in \mathbb{U}, \quad Q(z) \neq 0$$ For $t \in [-\pi, \pi]$, we set $$f(t) = F\left(e^{it}\right) = g(t) + ih(t) \quad \text{where} \quad (g(t), h(t)) \in \mathbf{R}^{2}$$ For $u \in [-\pi, \pi]$, we define a function $f_{u}$ from $[-\pi, \pi]$ to $\mathbf{R}$ by $$\forall t \in [-\pi, \pi], \quad f_{u}(t) = g(t)\cos(u) + h(t)\sin(u) = \operatorname{Re}\left(e^{-iu}F\left(e^{it}\right)\right) = \operatorname{Re}\left(e^{-iu}f(t)\right).$$
In this question, we fix $u \in [-\pi, \pi]$ and assume that $f_{u}$ is not constant. We also fix $y \in \mathbf{R}$. Using if necessary the expression of $f_{u}(t)$ as the real part of $e^{-iu}F\left(e^{it}\right)$ and Euler's formula for the real part, determine $S \in \mathbf{C}_{2n}[X]$ such that $$\forall t \in [-\pi, \pi], \quad f_{u}(t) = y \Longleftrightarrow S\left(e^{it}\right) = 0.$$ Deduce that the set $f_{u}^{-1}(\{y\}) \cap [-\pi, \pi[$ is finite with cardinality bounded by $2n$.