TITLE: Sum over the powers of the roots of unity $\sum \omega_j^k$
QUESTION [5 upvotes]: 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. $\hspace{6cm}\blacksquare$

REPLY [4 votes]: Yes, one can explain that the vector average of the vertices of a regular polygon is zero with geometry - in particular, with symmetry. If the average were nonzero it would be biased to one particular part of the polygon, which is impossible. Explicitly, the vertex set is rotation-invariant which implies their vector average is rotation-invariant, the only vector of which is zero. (Here the rotations we are considering are just rotations by integer multiples of $2\pi/n$.)
This reasoning can, of course, be phrased algebraically: if $\xi\ne0$ is a nontrivial root of unity then
$$\xi\sum \zeta^k=\sum \zeta^k \quad\implies\quad (\xi-1)\sum \zeta^k=0 \quad\implies\quad \sum \zeta^k=0.$$