How can we prove that the sum of the vertex coordinates of a regular polygon centered at $0$ is $0$?

Question

I'm trying to formally see this:

Let $a_1, \dots, a_n \in \mathbb{C}$ the vertexs of a regular polygon centered at $0$. Prove that $$\sum_{j=1}^{n}a_j=0.$$

I've tried by writing the vertex like this:

$$\sum_{j=1}^{n}re^{i\displaystyle\frac{2\pi(j-1)}{n}},$$

but I can't get nothing.

Any hint will be appreciated. Thanks for your time.

Answer 1

Let $\alpha = \exp\left(\frac{2\pi}n i\right)$, then

$$\sum_{j=1}^{n}r\exp\left(i\displaystyle{\frac{2\pi(j-1)}{n}}\right) = r\sum_{j=0}^{n-1} \alpha^j = r \frac{\alpha^n - 1}{\alpha -1} = r \cdot 0 = 0.$$

Answer 2

What happens when you multiply the sum by a primitive $n$th root of unity? All it does is permute the terms in the sum. Given $a\neq 1$, when is it possible that $ax=x$?

Answer 3

$\omega_j = e^{i \,2\pi(j-1) / n}$ are the $n^{th}$ roots of unity, which to say that they are the $n$ roots of the polynomial $z^n-1\,$. By Vieta's formulas, their sum is $\;\sum_{j=0}^{n-1} \omega_j = 0\,$.

---

[ EDIT ]  Alternative hint: the centroid (barycenter) of an arbitrary polygon is $\frac{1}{n}\sum a_j\,$. In the case of a regular polygon, the centroid coincides with the center of symmetry, so given that the polygon is centered at $0$ it follows that $\sum a_j = 0\,$.