Proving $\sum_{k=1}^{n}\cos\frac{2\pi k}{n}=0$ [duplicate]

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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?

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3 Answers

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Your sum is the sum of the real parts of the $n$th roots of the unity. That is, they are the roots of the polynomial $z^n-1$. The sum of the roots is equal to the coefficient of $z^{n-1}$ by Vieta's formulas, hence it is zero.

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Observe that$$\sum_{k=1}^n\cos(\frac{2\pi k}{n})=\operatorname{Re}\left(\sum_{k=1}^ne^{\frac{2\pi ki}{n}}\right).$$Consider the sum$$\sum_{k=1}^ne^{\frac{2\pi ki}{n}}=\sum_{k=1}^n(e^{\frac{2\pi i}{n}})^k$$and use the fact that$$1+\sum_{k=1}^n r^k =\frac{1-r^{n+1}}{1-r}$$to show it's zero.

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The cosines are the $x$-coordinates of points on the unit circle equally spaced around the origin. Since the average $x$ value is zero, it follows that the sum of the $x$ values is also zero.

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