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I want to prove that the below equation can be held.

$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$

Firstly I tried to check the equation with small values of $n$

$$ \text{As } n=2 $$

$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \right) + \cos\left(\frac{ 2 \pi \cdot 2 }{ 2 } \right) $$

$$ = \cos\left(\pi\right) + \cos\left(2 \pi\right) $$

$$ = -1+ 1 =0 ~~ \leftarrow~~ \text{Obvious} $$

But

$$ \text{As}~~ n=3 $$

$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 3 } \right) +\cos\left(\frac{ 2 \pi \cdot 2 }{ 3 } \right) + \cos\left(\frac{ 2 \pi \cdot 3 }{ 3 } \right) $$

$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + \cos\left( 2\pi \right) $$

$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + 1 =?$$

What formula(s) or property(s) can be used to prove the equation?