I'm having a problem with the second part of the question, which shows that the sum of the vectors equals zero. I came up with a geometric proof, but I wondered how to prove it by using complex numbers and maybe links to the first part of the question. I was thinking about the equally distributed roots of unity, but don't know how to move on.
Let's say one of those points was $1$ and let $\gamma$ be a primitive $n$th root of unity then those $n$ points are simply $\gamma^{i}$ as $i$ goes from $0$ to $n-1$ so you have to show $ 1 + \ldots + \gamma^{n-1} = 0$ which is to show that $\gamma$ is root of the polynomial $1 + \ldots + x^{n-1}$. For the first part not that all the $n$th roots of unity are solutions to $x^n - 1 = 0$ and conversely all solutions are $n$th roots of unity. Now divide $x^n - 1$ by $x - 1$.
The roots for the equation $z^n-1=0$ are given by $1,$ $\omega,$ $\omega^2,$ $ \dots,$ $\omega^{n-1},$ where $\omega$ is a primitve root of unity. Take $\omega=e^{\frac{2 \pi i}{n}}.$
We have that $$z^n-1=(z-1)(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1}).$$ Divide both sides of the above equation by $z-1.$ We have
$$\frac{z^n-1}{z-1}=\frac{(z-1)(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})}{z-1}.$$ See that $z^n-1=(z-1)(1+z+z^2+\dots+z^{n-1}).$
So, $$(z-1)(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=1+z+z^2+\dots+z^{n-1},$$ which was to be shown.
As for second part, if $A_1,A_2,\dots,A_n$ are $n$ points equally spaced around a circle of radius $r$ centered at $O,$ so that they form vertices of a regular polygon of $n-$sides, then we have that $\vec{OA_n}=\omega^{n-1}.$
This means that $$\vec{OA_1}+\vec{OA_2}+\dots+\vec{OA_n}= 1+ \omega + \omega^2 + \dots + \omega^{n-1}= \text{Sum of all $\textit{n}$ $n-$th roots of unity}.$$ But we know that sum of all roots of a $n-$degree polynomial, $p(x)=a_nx^n+a_{n-1}x^{n-1}+ \dots+ a_0,$ is given by $-\frac{a_{n-1}}{a_n}.$
So, we have that $\text{Sum of all n roots of unity}=- \frac{0}{1}=0.$ Therefore, $$\vec{OA_1}+\vec{OA_2}+\dots+\vec{OA_n}=0.$$
