5

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

Case z^3=1

n=5

Case z^5=1

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. $\hspace{6cm}\blacksquare$

N3buchadnezzar
  • 10,633
  • 6
  • 44
  • 89

2 Answers2

4

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

$$\xi\sum \zeta^k=\sum \zeta^k \quad\implies\quad (\xi-1)\sum \zeta^k=0 \quad\implies\quad \sum \zeta^k=0.$$

anon
  • 80,883
  • 8
  • 148
  • 244
1

If $\gcd(n, k) = d$, then exponentiation by $k$ will give $d$ full sets of all $\frac{n}{d}$-th roots of unity. If $d = n$, these sums up to $n$. If not, then each of the $d$ sets will sum up to $0$.

Arthur
  • 187,016
  • 14
  • 158
  • 288