Show that for the $n$th roots of unity $w_0, ..., w_{n-1}$ where $n \gt 1$ that ... Show that for the $n$th roots of unity $w_0, ..., w_n$ where $n \gt 1$ that
$\sum _{j=0}^{n-1}w_j^k = 0$ for $1 \le k \le n-1$.
The book is suggesting that I use: $1 + z + z^2 + ... + z^n = \frac{1-z^{n+1}}{1-z}$ and let $z = w_1^k$ but I'm can't figure any relationship between the two.
Anyone have any ideas?
 A: Two corrections:


*

*Where you wrote $w_0, ..., w_n,$ you should have written $w_0, ..., w_{n-1}.$

*More importantly, the formula you wrote is not true for $k=0.$  The inequality should be $1 \le k \le n-1.$  (You can see that the sum is equal to $n$ if $k=0.)$
With those things said, if you use the fact that $$\omega_k=e^{2\pi ki/n},$$
you can see that the sum you want is the sum of a geometric progression, and you can apply the suggestion from your book.  (I've written $\omega,$ a lower-case Greek letter omega, instead of the letter $w,$ as you wrote it, since that's the common notation for this.)
A: See that $\sum_{j=0}^{n-1}w_j^k$ is really the same thing as $\sum_{j=0}^{n-1}w_j$ since $w_j^k$ is equivalent to some other root of unity $w_p$.
See that $w_j=w_1^j$ thanks to Demoivre's formula.
It then becomes a geometric sum of the form $\sum_{j=0}^{n-1}z^j=\frac{1-z^n}{1-z}$ where $z=w_1$.
See then that $1-w_1^n=1-1=0$ (recall what a root of unity is).
Thus, $\sum_{j=0}^{n-1}w_j^k=0$ for any whole number $1\le k\le n-1$.  (a small note is that this is not true for non-integer $k$, nor is it true for $k=0$, which is obvious.)
