$\sum_{j=0}^{n-1}z_j^k=\begin{cases} 0, & \text{if $1\leq k \leq n-1$ } \\ n, & \text{if $k=n$ } \end{cases}$

Question

Show that $\sum_{j=0}^{n-1}z_j^k=\begin{cases} 0, & \text{if $1\leq k \leq n-1$ } \\ n, & \text{if $k=n$ } \end{cases}$, where $z_0,...,z_{n-1}$ are the $n$-th roots of unity.

For $k=n$ it is trivial as all the numbers will be $1$. Hence we get $1+\cdots +1$ ($n$ times). But is there any quick process for the rest?

Answer 1

If $z_0$ is a primitve $n$-th root of unity, then $1, z_0^2,\ldots z_0^{n-1}$ are the remaining roots of unity.

So,

$$\sum_{j=0}^{n-1}z_j^k=\sum_{j=0}^{n-1}z_0^{kj}.$$

If $k=n$ this sum becomes $$\sum_{j=0}^{n-1}1=n.$$

If $1\le k\le n$, then $z_0^k$ is an $n$-th root of unity, and so the sum vanishes as in Olivier Ochoa's answer.

Answer 2

Hint. You may use the identity

$$ 1+z+z^2+\cdots+z^{n-1}=\frac{1-z^n}{1-z},\qquad z\neq 1, \tag1 $$

with $z:=z_j$ such that $z_j^n=1$ and $z_j \neq1$.

You may prove $(1)$ by evaluating $$ (1-z)\left(1+z+z^2+\cdots+z^{n-1}\right). $$ The case $z_j^n=1$ is direct.

Answer 3

Let $\omega = e^{\frac{2\pi}{n}}$. Then for $i = 0,1,\ldots,n-1$ we have $z_i = \omega^i$.

Let's just look at the case $k \neq n$.

If $\gcd(k,n) = 1$ then it's easy because raising every $n$-th root of unity to the $k$-th power is just permuting them so the sum remains the same as in the case of $k = 1$ (which is $0$).

If $\gcd(k,n) = d >1$ then raising every $n$-th root of unity to the $k$-th power creates $d$ identical sums

$$(1+\omega^d+\ldots+\omega^{n/d-1})+(1+\omega^d+\ldots+\omega^{n/d-1})+\ldots+(1+\omega^d+\ldots+\omega^{n/d-1})$$

You'll find that each sum consists of a full set of the $n/d$-th roots of unity, and so is equal to $0$.