I am trying to find the eigenvalue of cycle graph and its complement.
How to simplify.
Suppose $\omega^{1}+\omega^{n-1}=2\cos (2\pi/n) $, then, $\omega^{\frac{n-1}{2}}+\omega^{\frac{n+1}{2}}=\ ?$
Is it true that $\omega^{1}+\omega^{n-1}=\omega^{\frac{n-1}{2}}+\omega^{\frac{n+1}{2}},$ where $\omega=e^{\frac{2\pi i}{n}}$?
Similarly, what is the value of $\omega^{0}+0+\omega^{2}+\omega^{3}+\cdots+\omega^{n-2}+0$.
Both the adjacency matrices $C_n$ and its complement $\bar{C_{n}}$ gives circulant matrix.
We have $\omega$ is a primitive $n$th root of unity, $\omega^{k}$ and $\omega^{n-k}$ are conjugates. This means that $(1/2)(\omega^{k}+\omega^{n-k}) = \mathrm{Re}(\omega^{k}) = \cos{\frac{2\pi k}{n}}$ (and so $\omega^{\frac{n-1}{2}} + \omega^{\frac{n+1}{2}} = 2 \cos{\frac{(n-1)\pi}{n}}$).
Also, $\omega$ is a root of $$z^n -1 = (z-1)(z^{n-1} + \ldots + z+1),$$ so $$\omega^{n-1}+ \ldots + \omega+1 = 0,$$ therefore $$1 + 0 + \omega^{2} + \ldots + \omega^{n-2} + 0 = 0-\omega - \omega^{n-1} = -2\cos{\frac{2\pi}{n}}.$$