A circle with center at O is divided into n equal arcs, n≥ 2, by the
points A₁A₂ A₃....A_{n} Find the sum of radius vectors i OA_{i}
where
, i=1,2,...,n.
Let's use the circle whose radius is $\displaystyle 1$ in the complex plane. Assume that $\displaystyle A_{0}=1$.
Your points are the $\displaystyle n^{th}$ roots of the unity. Let $\displaystyle \omega$ be a $\displaystyle n^{th}$ primitive root of the unity (i.e. $\displaystyle \omega^{n}=1$ and $\displaystyle \forall d<n, \omega^{d}\neq 1$).
Then your sum is $\displaystyle 1+\omega +...+\omega^{n-1}=0$
If you want to generalize this to your problem, just apply a homothety and a rotation to the circle we used to obtain yours.
Hello, makenqau!
A circle with center at $\displaystyle O$ is divided into $\displaystyle n$ equal arcs, $\displaystyle n \geq 2$
by the points: $\displaystyle A_1, A_2, A_3, \hdots A_n.$
Find the sum of radius vectors: .$\displaystyle \sum^n_{i=1} \overrightarrow{OA}_i $
This is a classic problem . . .
I recall that the sum is zero, but the proof eludes me.
With that in mind, you may be able to construct your own proof.