Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda \in \mathbb{C}$, $P_\lambda(X) = P(\lambda X) - P(\lambda)$. For all $\lambda$ in $\mathbb{C}$, $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X - 1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$. Using formula (I.1), show that $$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n} + 1}{X - \omega_k} \omega_k$$ then deduce that $$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1 - \omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1 - \omega_k)^2}.$$
Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda \in \mathbb{C}$, $P_\lambda(X) = P(\lambda X) - P(\lambda)$. For all $\lambda$ in $\mathbb{C}$, $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X - 1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$.
Using formula (I.1), show that
$$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n} + 1}{X - \omega_k} \omega_k$$
then deduce that
$$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1 - \omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1 - \omega_k)^2}.$$