Adjacency spectra of path graph In Spectra of Graphs by Brouwer and Haemers it is written:

The ordinary spectrum follows by looking at $C_{2n+2}$. If $u(\zeta) = (1, \zeta, \zeta^2,...,\zeta^{2n+1})^T$ is an eigenvector of $C_{2n+2}$,
where $\zeta^{2n+2} = 1$, then $u(\zeta)$ and $u(\zeta ^{−1})$ have
the same eigenvalue $2\cos(\pi j/(n + 1))$, and hence so has $u(\zeta) − u(\zeta ^{−1})$. This latter vector has two zero coordinates
distance $n + 1$ apart and (for $\zeta \not= \pm 1$) induces an
eigenvector on the two paths obtained by removing the two points where
it is zero.

I do not understand two things:
1/ Why do $u(\zeta)$ and $u(\zeta ^{−1})$ have the same eigenvalue $2\cos(\pi j/(n + 1))$?
2/ Why is it that "this latter vector has two zero coordinates
distance $n + 1$ apart and (for $\zeta \not= \pm 1$) induces an
eigenvector"?
 A: \begin{align*}& z^{2n+2}=1\\\Rightarrow & z=e^{\frac{2k\pi\mathtt{i}}{2n+2}}=e^{\frac{k\pi\mathtt{i}}{n+1}},\text{ for }k=0, 1, \ldots, 2n+1\\\Rightarrow & \zeta=e^{\frac{\pi\mathtt{i}}{n+1}}.\end{align*}
Therefore, $$u(\zeta)=(1, \zeta, \zeta^2, \ldots, \zeta^n, -1, -\zeta, -\zeta^2, \ldots, -\zeta^n)$$ and
$$u(\zeta^{-1})=(1, \zeta^{-1}, \zeta^{-2},\ldots, \zeta^{-n},-1, -\zeta^{-1}, -\zeta^{-2}, \ldots, \zeta^{-n}).$$
Now here from I think you can get your answer.
A: The quoted part of the textbook is slightly incorrect: depending on which $(2n+2)$th root of unity $\zeta$ we choose, we get a particular eigenvalue of $u(\zeta)$ (and $u(\zeta^{-1})$).
Precisely, let $\zeta = e^\frac{2\pi i j}{2n+2}$ be a $(2n+2)$th root of unity for a fixed $j \in \{0,1,\dotsc,2n+1\}$. Then, the vectors $u(\zeta)$ and $u(\zeta^{-1})$ are eigenvectors of $C_{2n+2}$ having the common eigenvalue $2 \cos(\pi j/(n+1) )$.
One can see this by following a similar argument as in @G_0_pi_i_e's answer, which explains the scenario when $j = 1$.

Note that in the case $j = 0$ we get $u(\zeta) = (1,1,\dotsc,1) = u(\zeta^{-1})$, and in the case $j = 2n+1$ we get $u(\zeta) = (-1,1,\dotsc,-1,1) = u(\zeta^{-1})$. In either case, $u(\zeta) - u(\zeta^{-1})$ is the zero vector, so no eigenvector is induced on $P_n$ (recall that an eigenvector must be a nonzero vector by definition).
Furthermore, note that this does not happen for any other value of $j$ in this set, because if $u(\zeta) = u(\zeta^{-1})$, then $\zeta = \zeta^{-1}$, so that $\zeta^2 = 1$. Thus, either $\zeta = 1$ or $\zeta = -1$, that is, $j = 0$ or $j = 2n+1$. Hence, we do get eigenvectors of $P_n$ for the other values of $j$.
Finally, since $\zeta^{-1} = \zeta^{2n-1}$ for any $(2n+2)$th root of unity $\zeta$, we see that $u(\zeta) - u(\zeta^{-1}) = u(\zeta'^{-1}) - u(\zeta')$ when $\zeta = e^\frac{2\pi i j}{2n+2}$ and $\zeta' = e^\frac{2\pi i (2n+2-j)}{2n+2}$. Hence, the distinct eigenvalues of $P_n$ are given by $2\cos(\pi j /(n+1))$, $j = 1,2,\dotsc,n+1$.
