Question: Recently I came across this summation:
$$\sum\_{k=0}^n\cos\Bigl(k\frac{2\pi}{n}\Bigr)$$ when I was trying to evaluate the following summation $$\sum\_{k=0}^{n} z^k+z^{-k}$$ where $z \in \mathbb{C}, z=\operatorname{cis}(\frac{2\pi}{n})$

But I got stuck... Could anyone give me a tip ?

Thanks in advance!

Answer: 
$\sum\limits\_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$

$\sum\limits\_{k=0}^n z^{-k}=z\frac{1-z^{-(n+1)}}{z-1}$

Is that where you started.
**Hint**:

$\cos\dfrac{2k\pi}n$ is the real part of $\:\Bigl(\mathrm e^{\tfrac{2i\pi}n}\Bigr)^{\!k}$.