The roots of unity are defined as the points $\omega_n$ to satisfy
$$
z^n = 1
$$
Explicitly these can be given as
$$
\omega_k = \exp(2i\pi k ) \,,\qquad 1 \leq k \leq n-1
$$
In my book it is stated that
$$
\sum_{j=0}^{n-1} \omega_j^k =
\left\{
\begin{array}{ll}
0 \, , & 1 \leq k \leq n - 1 \\
n \, , & k = n
\end{array}
\right.
\tag{1}
$$
and I have proven this *algebraically* see below. However I want to look at the problem
from a geometrical perspective. As long as $k$ and $n$ are coprime, eg $\gcd(n,k)=1$
then the powers of $\omega^k$ seem to simply *permutate* $\omega$.

**n=3**

**n=5**

Can this property of the roots of units be explained geometrically, or by some combinatorial proof? My question is

**Can equation $(1)$ be explained purely from a geometrical or combinatorial point of view?**

Proof: Assume that $1\leq k \leq n-1$ then \begin{align*} \sum_{j=0}^{n-1} w_j^k & = \sum_{j=0}^{n-1} [\exp(2\pi i k/n)]^j = \frac{1 - [\exp(2\pi i k/n)]^n}{1 - \exp(2\pi i k/n) } = 0 \end{align*} Since the denominator equals zero, eg $\exp(2i\pi k)=1\ \forall k\in \mathbb{N}$. The case for $k=n$ must be treated seperately as it leads to a $0/0$ expression above. \begin{align*} \sum_{j=0}^{n-1} w_j^n & = \sum_{j=0}^{n-1} \exp(2\pi i j) = n \end{align*} Where again $\exp(2\pi i k)=1$. This finishes the proof. $\hspace{6cm}\blacksquare$