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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 $u(\zeta)$ and $u(\zeta ^{−1})$ have the same eigenvalue $2\cos(\pi j/(n + 1))$?

2/ Why holds that "this latter vector has two zero coordinates distance $n + 1$ apart and (for $\zeta \not= \pm 1$) induces an eigenvector"?

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    You are probably trying to find out the spectrum of the path graph on $n$ vertices. Right! – 2017-01-09

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\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.