Suppose that we have $n$ vectors $v_1,v_2...v_n$ with same magnitude in plane s.t. the angle between $v_i$ and $v_{i+1}$ is $2\pi/n$ then $v_1+v_2+...v_n=0$ for all $n \geq 2$.
I can show this by using complex numbers, but I wonder whether it has more elementary way to show it. (At least without using complex numbers)
Yes, there is an elementary and somewhat an elegant solution. Denote $$v=v_1+\dots+v_n.$$ We're going to use the $v_i$'s as vectors on the plane, $v_i=\overset{\to}{OV_i}$. If we rotate the whole plane(or just that vectors) with respect to point $O$ and by the angle $2\pi/n$, then the sum of $v_i$'s doesn't chenge (since the angles between them is $2\pi/n$), but the sum $v$ changes if it's not the zero-vector.