This is inspired by another problem involving the seventh roots of unity and one of the comments there. I want to prove that if $\omega$ is the seventh roots of unity, then $(\omega^{n+1}+\omega^{n+2}+\omega^{n+4}), n\in\mathbb{Z}$ consists of the vertices of a regular heptagon which lies on a circle with radius $\sqrt{2}$
My solution:
$(\omega+\omega^{2}+\omega^{4})(\omega^{3}+\omega^{5}+\omega^{6})=2$.
Furthermore, $\omega^{3}+\omega^{5}+\omega^{6}$ is the complex conjugate of $\omega+\omega^{2}+\omega^{4}$. Therefore $\omega+\omega^{2}+\omega^{4}$ is one of the points on a circle with radius $\sqrt{2}$: $z\bar{z}=2$.
$\omega^{n+1}+\omega^{n+2}+\omega^{n+4}$ is $\omega+\omega^{2}+\omega^{4}$ rotated by $\frac{2\pi n}{7}$.
I am wondering if there are other solutions for this one
There is an even simpler way. Let :
$$\alpha_n:=\omega^{n+1}+\omega^{n+2}+\omega^{n+4}\tag{1}$$
We can evidently write :
$$\alpha_{n+1}=\omega \alpha_n\tag{2}$$
The geometrical interpretation of (2) is as follows:
$\alpha_{n+1}$ is obtained as the image of $\alpha_n$ by the $\frac{2\pi}{7}$ rotation associated with multiplication by $\omega$ !
Please note that this reasoning can be applied to any combination of the form $\alpha_n:=a\omega^n+b\omega^{n+1}+...$.