A monic polynomial of degree $d \geq 1$ $$P = \sum _ { i = 0 } ^ { d } a _ { i } X ^ { i } \in \mathbb { C } [ X ]$$ is called reciprocal if $a _ { i } = a _ { d - i }$ for $0 \leq i \leq d$.
(a) Show that a monic polynomial $P \in \mathbb { C } [ X ]$ of degree $d$ is reciprocal if and only if $X ^ { d } P \left( \frac { 1 } { X } \right) = P$.
(b) Let $P \in \mathbb { C } [ X ]$ be a monic reciprocal polynomial. Show that if $x \in \mathbb { C }$ is a root of $P$, then $x \neq 0$ and $\frac { 1 } { x }$ is also a root of $P$, with the same multiplicity.

If $\alpha$ is an algebraic number with minimal polynomial $\Pi _ { \alpha }$, the complex roots of $\Pi _ { \alpha }$ different from $\alpha$ are called the conjugates of $\alpha$. We denote by $C ( \alpha )$ the set of conjugates of $\alpha$.
Let $x$ be an algebraic number of modulus 1 and such that $x \notin \{ - 1,1 \}$. Show that $\frac { 1 } { x }$ is a conjugate of $x$. Deduce that $\Pi _ { x }$ is reciprocal.

For every integer $n > 1$, we define $P _ { n } \in \mathbb { Z } [ X ]$ by $$P _ { n } = X ^ { 4 } - ( 6 + n ) X ^ { 3 } + ( 10 + n ) X ^ { 2 } - ( 6 + n ) X + 1 .$$ Show that if $x \in \mathbb { C }$ is a root of $P _ { n }$, then $\frac { 1 } { x }$ is also a root of $P _ { n }$, with the same multiplicity.

Let $P ( X )$ be a monic polynomial of degree $d \geqslant 1$ with complex coefficients which we write in the form: $$P ( X ) = a _ { 0 } + a _ { 1 } X + a _ { 2 } X ^ { 2 } + \cdots + a _ { d - 1 } X ^ { d - 1 } + X ^ { d }$$ We assume that $a _ { 0 } \neq 0$. We denote by $\lambda _ { 1 } , \ldots , \lambda _ { d } \in \mathbb { C }$ the roots of $P ( X )$ (with multiplicity). For all integers $n \geqslant 1$, we define: $$N _ { n } = \lambda _ { 1 } ^ { n } + \lambda _ { 2 } ^ { n } + \cdots + \lambda _ { d } ^ { n }$$
Let $Q ( X )$ be the reciprocal polynomial of $P ( X )$ defined by $Q ( X ) = X ^ { d } P \left( \frac { 1 } { X } \right)$. Show that: $$\begin{aligned} Q ( X ) & = 1 + a _ { d - 1 } X + \cdots + a _ { 1 } X ^ { d - 1 } + a _ { 0 } X ^ { d } \\ & = \left( 1 - \lambda _ { 1 } X \right) \left( 1 - \lambda _ { 2 } X \right) \cdots \left( 1 - \lambda _ { d } X \right) \end{aligned}$$

Let $P ( X )$ be a monic polynomial of degree $d \geqslant 1$ with complex coefficients which we write in the form: $$P ( X ) = a _ { 0 } + a _ { 1 } X + a _ { 2 } X ^ { 2 } + \cdots + a _ { d - 1 } X ^ { d - 1 } + X ^ { d }$$ We assume that $a _ { 0 } \neq 0$. We denote by $\lambda _ { 1 } , \ldots , \lambda _ { d } \in \mathbb { C }$ the roots of $P ( X )$ (with multiplicity). For all integers $n \geqslant 1$, we define: $$N _ { n } = \lambda _ { 1 } ^ { n } + \lambda _ { 2 } ^ { n } + \cdots + \lambda _ { d } ^ { n }$$
Let $Q ( X )$ be the reciprocal polynomial of $P ( X )$ defined by $Q ( X ) = X ^ { d } P \left( \frac { 1 } { X } \right)$. Show that: $$\begin{aligned} Q ( X ) & = 1 + a _ { d - 1 } X + \cdots + a _ { 1 } X ^ { d - 1 } + a _ { 0 } X ^ { d } \\ & = \left( 1 - \lambda _ { 1 } X \right) \left( 1 - \lambda _ { 2 } X \right) \cdots \left( 1 - \lambda _ { d } X \right) \end{aligned}$$

Let $P \in \mathbf{C}[X]$ of degree $p$. We write $P = \sum_{k=0}^{p} a_k X^k$, where $a_0, \ldots, a_p$ are complex numbers, and $a_p \neq 0$.
Show that $P$ is reciprocal if and only if for every integer $k$, $0 \leq k \leq p$, we have the equality $a_k = a_{p-k}$.

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 $m_i$.

Let $Q$ be a polynomial of degree $p$. We say that $Q$ is antireciprocal if $$Q(X) = -X^p Q\left(\frac{1}{X}\right)$$ Show that if $Q$ is antireciprocal, 1 is a root of $Q$ and there exists a polynomial $P$ that is constant or reciprocal such that $Q = (X-1)P$.

Let $R$ be a non-constant polynomial of $\mathbf{C}[X]$ having the following property: Every root $a$ of $R$ is nonzero and $\frac{1}{a}$ is a root of $R$ with the same multiplicity as $a$.
Prove that the product of the roots of $R$, counted with multiplicities, can only take the values 1 or $-1$. One may note that the equality $a = \frac{1}{a}$ holds only for $a = 1$ or $-1$.

Let $R$ be a non-constant polynomial of $\mathbf{C}[X]$ having the following property: Every root $a$ of $R$ is nonzero and $\frac{1}{a}$ is a root of $R$ with the same multiplicity as $a$.
Deduce that $R$ is reciprocal or antireciprocal.

Let $P \in \mathbf{C}[X]$ of degree $p$. We write $P = \sum_{k=0}^{p} a_k X^k$, where $a_0, \ldots, a_p$ are complex numbers, and $a_p \neq 0$. Show that $P$ is reciprocal if and only if for every integer $k$, $0 \leq k \leq p$, we have the equality $a_k = a_{p-k}$.

Let $Q$ be a polynomial of degree $p$. We say that $Q$ is antireciprocal if $$Q(X) = -X^p Q\left(\frac{1}{X}\right)$$ Show that if $Q$ is antireciprocal, 1 is a root of $Q$ and that there exists a polynomial $P$ that is constant or reciprocal such that $Q = (X-1)P$.

Let $R$ be a non-constant polynomial of $\mathbf{C}[X]$ having the following property: Every root $a$ of $R$ is nonzero and $\frac{1}{a}$ is a root of $R$ with the same multiplicity as $a$. Prove that the product of the roots of $R$, counted with multiplicities, can only take the values 1 or $-1$. One may note that the equality $a = \frac{1}{a}$ holds only for $a = 1$ or $-1$.

Let $R$ be a non-constant polynomial of $\mathbf{C}[X]$ having the following property: Every root $a$ of $R$ is nonzero and $\frac{1}{a}$ is a root of $R$ with the same multiplicity as $a$. Deduce that $R$ is reciprocal or antireciprocal.

Until the end of part A, we assume that all roots of $p$ are stable and have multiplicity 1.
Justify that there exists $\lambda \in \{-1, 1\}$ such that $p = \lambda p_0$.