Geometric Interpretation of a question related to complex numbers..

Question

QUESTION: I just encountered that $$\sum_{n=0}^8e^{in\theta}=0$$where $\theta=\frac{2π}9$

First I prove it, and then ask my question :)..

($i=\sqrt{-1}$)

---

MY ANSWER: We know, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ and that $e^{in\theta}=\cos(n\theta)+i\sin(n\theta)$.. but writing that out in this summation really makes it nasty.. let's consider the series-

$1+e^{i\theta}+(e^{i\theta})^2+......+(e^{i\theta})^8$

If we use g.p. we get the summation is $$\frac{(e^{i\theta})^9-1}{e^{i\theta}-1}$$

Now, $$e^{9i\theta}=\cos(9\theta)+i\sin(9\theta)=1$$ considering $\theta=\frac{2π}9$ as given.

Therefore the summation is zero..

Q.E.D. $\square$

---

Now, if we observe we find out that for any $\theta=\frac{2π}k$ we have $$\sum_{n=0}^{k-1}e^{in\theta}=0$$

Isn't that great ?

Forgive me if this looks silly to you :P, but I was so elated after finding this out..

My question is, since almost every concept involving complex numbers can be chalked out geometrically.

What is the geometric interpretation of this pattern we discover?

Answer 1

We can think of $e^{in\theta}$ as a vector $v_n$ from $0$ which ends at $e^{in\theta}$.

Now, let $v=v_0+v_1+...v_8$, then rotation around origin for $\theta $ takes $v$ to it self. But the only vector which non trivial rotation fixes is zero vector and thus $v=0$.

Answer 2

A regular polygon's centre of mass if each of its vertices has the same weight, while the edges are weightless. (A uniform density on the edges gets the same centre of mass, but the easiest way to prove this builds on the weightless edges case.)