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I have this question

Let $n \ge 2$ be an integer. Prove that $$\sum_{k=0}^{n-1}\cos(\frac{2k\pi}{n}) = 0 = \sum_{k=0}^{n-1}\sin(\frac{2k\pi}{n})$$

I was given the hint to

Set $z = \cos(\frac{2\pi}{n}) + i\sin(\frac{2\pi}{n})$, so $z^n=1$. Now write this as $(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)=0$ and go from there...

I get that $(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)=0$ is just an expansion and rearrangement of $z^n=1$, but what I don't understand is, why set $z = \cos(\frac{2\pi}n) + i\sin(\frac{2\pi}{n})$ and how do we know that $z^n=1$

Parcly Taxel
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Alex Twin
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  • We set $z= \cos(\frac{2\pi}{n}) + i\sin(\frac{2\pi}{n})$, so obviously $z=e^{2 \pi i/n }$ which implies that $z^n=(e^{2 \pi i/n })^n=e^{2 \pi i}=1.$ – Aleph-null Dec 06 '17 at 04:22
  • Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 22 '18 at 21:36

2 Answers2

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HINT Let’s take a look here, it’s about: https://en.m.wikipedia.org/wiki/Root_of_unity

As a graphical interpretation it is the sum of the vectors lying on a unitary circle on the vertex of a regular polygon, thus they sum up to 0 by symmetry.

user
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When you set $z:=\text{cis}\left( \dfrac{2\pi}{n} \right)$ you win all the solutions to $z^n-1=0$, in factorization, using one of moivre law's ($\text{cis}^m(x)=\text{cis}(m.x)$),

$$ (z-1)\cdot \sum_{k=0}^{n-1} \text{cis}\left( \dfrac{2\pi}{n} \cdot k\right) = 0 + 0.\text{i} $$

If $z \neq 1$, we have that crazy sum are $0$. But the real part of this sum is $0$ and the imaginary part also is $0$.

The real part of $\displaystyle \sum_{k=0}^{n-1} \text{cis}\left( \dfrac{2\pi}{n} \cdot k\right) $ are $\displaystyle \sum_{k=0}^{n-1} \cos\left( \dfrac{2\pi}{n} \cdot k\right) = 0$ and imaginary part are $\displaystyle \sum_{k=0}^{n-1} \sin\left( \dfrac{2\pi}{n} \cdot k\right) = 0$. $\Box$

Gustavo Mezzovilla
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