Roots of unity are the solutions of the complex polynomial $t^{n}-1=0$ they have the following form $E_{n}=\{e^{\frac{2\pi ik}{n}}:k\in\mathbb{Z}\}=\{e^{\frac{2\pi ik}{n}}:k=1,...,n-1\}$. From the properties of the $e$-function we know that $|e^{\frac{2\pi ik}{n}}|=1$ for all roots of unity, hence they lie on the unit circle. They are eventually used to construct regular n-gon's.
Now, everywhere I read about them, it's stated that they are evenly spaced around the unit circle. However I'd like to know how to prove this.
As pointed out, the roots of
$t^n - 1 = 0 \tag 1$
are the $n$ complex numbers
$\omega^j = e^{2\pi i j / n} = (e^{2\pi i / n})^j, \; 0 \le j \le n - 1. \tag 2$
If we use the Euler identity on the $\omega^j$ we find
$\omega^j = e^{2\pi i j / n} = \cos \dfrac{2\pi j}{n} + i \sin \dfrac{2\pi j}{n}; \tag 3$
it is easy to see from $(3)$ that the ray emanating from the origin and passing through $\omega^j$ makes an angle $2\pi j / n$ with the positive $x$-axis; thus the angle between consecutive roots of unity $\omega^j$ and $\omega^{j + 1}$ is precisely $2\pi /n,$ no matter what the value of $j$; it is the same for any two consecutive $n$-th roots of unity, so the arc subtended by the angle 'twixt two consecutive $\omega^j$ always is of length $2 \pi / n$; they are evenly space around the unit circle.