For $\alpha \in \mathbb{C}$, we set $P_\alpha = X^2 + \alpha$. We denote by $G$ the set of complex polynomials of degree 1, and the inverse of $U \in G$ under composition is denoted $U^{-1}$.
Let $P$ be a complex polynomial of degree 2. Justify the existence and uniqueness of $U \in G$ and $\alpha \in \mathbb{C}$ such that $U \circ P \circ U^{-1} = P_\alpha$. Determine these two elements when $P = T_2$.

For $\alpha \in \mathbb{C}$, we set $P_\alpha = X^2 + \alpha$. We denote by $G$ the set of complex polynomials of degree 1, and the inverse of $U \in G$ under composition is denoted $U^{-1}$. We seek families $(F_n)_{n \in \mathbb{N}}$ of complex polynomials satisfying $$\forall n \in \mathbb{N}, \quad \deg F_n = n \quad \text{and} \quad \forall (m,n) \in \mathbb{N}^2, \quad F_n \circ F_m = F_m \circ F_n \tag{III.1}$$
Deduce the Block and Thielmann theorem: if $(F_n)_{n \in \mathbb{N}}$ satisfies (III.1), then there exists $U \in G$ such that $$\forall n \in \mathbb{N}^*, \quad F_n = U^{-1} \circ X^n \circ U \quad \text{or} \quad \forall n \in \mathbb{N}^*, \quad F_n = U^{-1} \circ T_n \circ U$$

We choose an even polynomial in $B_{N}$ (see question 2(c)), and we denote it $R_{N}$.
Show that there exist non-negative integers $r, s, t \geqslant 0$, real numbers $c_{1}, \ldots, c_{r}$ different from $\pm 1$, non-zero reals $\rho_{1}, \ldots, \rho_{s}$ and complex numbers $w_{1}, \ldots, w_{t}$ that are neither real nor purely imaginary, such that $$R_{N}(X) = \prod_{j=1}^{r} \frac{X^{2} - c_{j}^{2}}{1 - c_{j}^{2}} \prod_{k=1}^{s} \frac{X^{2} + \rho_{k}^{2}}{1 + \rho_{k}^{2}} \prod_{\ell=1}^{t} \frac{X^{2} - w_{\ell}^{2}}{1 - w_{\ell}^{2}} \cdot \frac{X^{2} - \overline{w_{\ell}}^{2}}{1 - \overline{w_{\ell}}^{2}}.$$

Let $n \in \mathbb{N}$ and let $P_n \in \mathbb{R}[X]$ be such that, for all $\theta \in \mathbb{R}$, $\sin((2n+1)\theta) = \sin(\theta) P_n\left(\sin^2(\theta)\right)$.
Determine the roots of $P_n$ and deduce that, for all $x \in \mathbb{R}$, $$P_n(x) = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{x}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$

Let $n \in \mathbb{N}^*$ and $$T _ { n } ( X ) = \sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } ( - 1 ) ^ { p } \binom { n } { 2 p } X ^ { n - 2 p } \left( 1 - X ^ { 2 } \right) ^ { p }.$$ For $k \in \llbracket 1 , n \rrbracket$, we set $y _ { k , n } = \cos \left( \frac { ( 2 k - 1 ) \pi } { 2 n } \right)$. Show that $$T _ { n } ( X ) = 2 ^ { n - 1 } \prod _ { k = 1 } ^ { n } \left( X - y _ { k , n } \right).$$

Let $P$ be a polynomial of degree $p$ written in factored form $P = a_p \prod_{i=1}^{d} (X - \lambda_i)^{m_i}$, where $\lambda_1, \ldots, \lambda_d$ are the distinct complex roots of $P$ and $m_1, \ldots, m_d$ their multiplicities. Write in factored form the polynomial $X^p P\left(\frac{1}{X}\right)$ and prove that if $P$ is reciprocal then for every integer $i$, $1 \leq i \leq d$, $\lambda_i$ is nonzero and $\frac{1}{\lambda_i}$ is a root of $P$ with multiplicity $m_i$.

Show that $p_0$, the reciprocal polynomial of $p$, satisfies $$\forall x \in \mathbf{R}^* \quad p_0(x) = x^n p(1/x)$$ and deduce that $$p_0 = a_n \prod_{j=1}^{n} \left(1 - \alpha_j X\right)$$