Solve $e^{ina}=-1$

Question

Solve the following equation: $$e^{ina}=-1 \quad a\in\mathbb{R},\ n \in \mathbb{N}^{*}$$

My thoughts:

\begin{align*} e^{ina}&=-1\\ &=e^{i\pi} \\ \iff na&=\pi [2\pi] \\ \end{align*} but why they wrote: $$na\equiv n\pi [2\pi] $$

Answer 1

My French is mediocre, however, the book is correct. It is saying that $-1$ is a root of the polynomial $P_n$ if and only if $na \equiv n\pi \; [2\pi]$. This is true because \begin{align*} -1 \text{ is a root of } P_n(z) &\iff w = (-1)^n \text{ satisfies } w^2 - 2 \cos(na) z + 1 \\ &\iff \text{For } w = (-1)^n = e^{in\pi}, w = e^{ina} \text{ or } w = e^{-ina}\\ &\iff e^{in\pi} = e^{ina} \text{ or } e^{in\pi} = e^{-ina} \\ &\iff na \equiv n\pi \;[2\pi] \text{ or } na \equiv -n\pi \;[2\pi] \end{align*} But now, $n\pi \equiv -n\pi \; [2\pi]$. So the two conditions are the same, and we have $$ -1 \text{ is a root of } P_n(z) \iff na \equiv n \pi \; [2\pi]. $$