Roots of unity - sum over all the solutions is zero?

Question

Show that for all roots of unity $z\neq 1$, meaning solutions of the equation $z^n=1$ with $n\in \mathbb{N}$ the following is true:

$\sum_{k=0}^{n-1}z^k=0$.

I'm totally lost here. I thought I would do this by induction, but I can't even find a solution to the equation besides 1.

Answer 1

Note That if $z=e^{2 \pi i/n}$ is a root, we have $z^n=1$

We have the following equality: $z^0 + z + z^2 + \ldots + z^{n-1} = \frac{z^{n}-1}{z-1} = 0$ because $z^n-1 = 1-1 = 0$

Answer 2

we have

$z(1+z+z^2+....z^{n-1})=z.S$

$=z+z^2+.......+z^{n-1}+1=S$

thus $S=zS$ and $S=0$ since $z\neq 0$ and $z\neq 1.$