Arguments and complex roots of unity

Question

I am confused about the following: $e^{\frac{2 \pi k i}{n}}$ and $e^{\frac{(\theta + 2 \pi k)i}{n}}$, is the difference between them that one starts from zero angles and the other was rotated by an angle of $\theta$? I was watching a lecture on roots, among other things, (see here: https://youtu.be/0p9y6u0VyVk?t=1821 ) and when I wanted to prove that the sum of the $n$ nth roots of unity is zero, using geometric series I realized I didn't understand something. $S = \frac{1-w^{n}}{1-w},$ where $w^{n}=e^{\frac{2 \pi i n}{n}}= e^{2 \pi i} = 1$ because it rotates back to angle zero. Honestly I am not sure what I am missing or not understanding here. Any help appreciated

Answer 1

For a complex number $z$, you can think of multiplication by $e^{i\theta}$ for $\theta\in[0,2\pi)$ as a rotation counterclockwise about the origin by an angle of $\theta$; that is, $ze^{i\theta}$ is the rotated version of $z$. Let $w=e^{2\pi i/n}$, then if we want to find the sum $$S=1+e^{2\pi i/n}+e^{2\times 2\pi i/n}+\cdots+e^{(n-1)2\pi i/n},$$ visualise this as the sum of the $n$ points on the unit circle in the complex plane spaced an angle of $2\pi/n$ apart. Now if we multiply $S$ by $e^{\pi i/n}$, this rotates all the points counterclockwise by an angle of $2\pi/n$, which just gives us back the original sum. So $e^{2\pi i/n}S=S$. But how can the rotated version of a complex number $S$ be itself? Well, it has to be $0$.