How to use factorisation $z^{n}-1$ to prove the summation goes to zero?

Question

I need to show that $\sin(\frac{2\pi}{n})+\sin(\frac{4\pi}{n})+...+\sin(\frac{2(n-1)\pi}{n})=0$ for $n>1$ but specifically using the fact that $z^{n}-1=(z-1)(1+z+z^{2}+...+z^{n-1})$ for every $z\in \mathbb{C}$ and $n\in \mathbb{N}$. I've proved that equality by induction but I do not know how to incorporate it what I need to show. Thinking it might have something to do with $n^{th}$ roots of unity.

Anything would be helpful. Thanks!

Answer 1

$\sin(\frac{2\pi}{n})+...+\sin(\frac{2(n-1)\pi}{n})$ is the imaginary part of $1+\zeta+\zeta^2+...+\zeta^{n-1}$ where $\zeta=e^{2\pi i/n}$.

$\zeta-1\ne 0$, but $\zeta^n=1$, so using the factorization you get the desired result.

Answer 2

Note $$ I= 1+e^{i \frac{2\pi}{n}} + e^{i \frac{4\pi}{n}}+ ... + e^{i \frac{2(n-1)\pi}{k}}= \frac {(e^{i \frac{2\pi}{n}})^n -1}{e^{i \frac{2\pi}{n}} -1} =0 $$

Thus, with $\sin x=\operatorname{Im }(e^{ix})$,

$$ \sin\frac{2\pi}{n}+ \sin\frac{4\pi}{n}+ ... + \sin\frac{2(n-1)\pi}{n}=\operatorname{Im} (I)=0 $$

Answer 3

Hints:

1. From the given factorisation, you deduce, not only the sum of the geometric series, but also, closely related, that $$z+z^2+\dots +z^{n-1}=z(1+z+\dots+z^{n-2})=\frac{z(z^{n-1}-1)}{z-1}.$$
2. Setting $z=\mathrm e^{\tfrac{2i\pi}n}$, $\sin(\frac{2\pi}{n})+\sin(\frac{4\pi}{n})+...+\sin(\frac{2(n-1)\pi}{n})$ is the imaginary part of $$z+z^2+\dots+z^{n-1}.$$