Evaluating $\frac{1}{N} \sum_{k=0}^{N-1}|1+\cos(\frac{2k\pi}{N})|^2$ for $N \ge 3, N \in \mathbb{N}$

Question

I've tried to write it in complex exponential form to see if I could spot any geometric series, which led me nowhere near a solution. Mathematica evaluates it to $\frac{3}{2}$, any idea on how to arrive to this result?

Answer 1

Write $\zeta=\exp(2\pi i/N)$. Then $$\cos\left(\frac{2\pi k}N\right)=\frac{\zeta^k+\zeta^{-k}}{2}$$ and then $$\left|1+\cos\left(\frac{2\pi k}N\right)\right|^2 =\left(1+\cos\left(\frac{2\pi k}N\right)\right)^2 =\frac{(\zeta^k+2+\zeta^{-k})^2}{4} =\frac{\zeta^{2k}+4\zeta^k+6+4\zeta^{-k}+\zeta^{-2k}}{4}.$$ Sum this from $k=0$ to $N-1$. The sums $\sum_{k=0}^{N-1}\zeta^{\pm 2k}$ and $\sum_{k=0}^{N-1}\zeta^{\pm k}$ are GPs which vanish (why?) leaving one with $$\sum_{k=0}^{N-1}\left|1+\cos\left(\frac{2\pi k}N\right)\right|^2=\frac{6N}{4}$$ etc.

Answer 2

Hint: do some trigonometry first: by the linearisation formulæ, $$\sum_{k=0}^{N-1}|1+\cos(\frac{2k\pi}{N})|^2 = \sum_{k=0}^{N-1}\Bigl(1+2\cos \frac{2k\pi}{N} +\cos^2 \frac{2k\pi}{N}\Bigr)=\sum_{k=0}^{N-1}\Bigl[1+2\cos \frac{2k\pi}{N} +\tfrac12\Bigl(1+\cos \frac{4k\pi}{N}\Bigr)\Bigr].$$

Now you have a formula for the sum of cosines (or sines) in arithmetic progression, which you may prove via the complex exponential: $$1 + \cos \theta +\cos 2 \theta + \dots + \cos n \theta = \frac{\sin \cfrac{(n +1) \theta }{2}}{\sin \cfrac{ \theta }{2}}\,\cos \frac{n \theta }{2} $$