Average of an odd number of equally spaced points on a circle I want to show that the average of an odd number of equally spaced points on the unit circle is equal to 0. More precisely, let $n$ be an odd number, $\theta_1,\ldots, \theta_n\in[0,2\pi)$ and $$
re^{i\psi}:=\frac{1}{n}\sum_{i=1}^ne^{i\theta_i}.$$ We want to show that if the $\theta_i$s are equally spaced then $r=0$. I remark that I do not want to use Vieta's formulas but rather prove it "by hand".
I was able to prove this formula for $r$:
$$r=\frac{1}{n}\left(n+2\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos(\theta_i-\theta_j)\right)^{1/2}
$$
(I wouldn't know if this is a well known formula or not...). If the angles are equally spaced, upon relabeling if necessary we have $\theta_i=\frac{2(i-1)\pi}{n},\:i=1,\ldots,n$ so that
\begin{align*}
r^2 & = \frac{1}{n^2}\left(n+2\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos(\theta_i-\theta_j)\right) \\
 & = \frac{1}{n^2}\left(n+2\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos\left(\frac{2(i-j)\pi}{n}\right)\right)
\end{align*}
Then we would like to show that
$$\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos\left(\frac{2(i-j)\pi}{n}\right)=-\frac{n}{2}.$$
If we set $n=2k+1\:k\in\mathbb{N}$ we can rewrite the previous formula in terms of $k$:
$$\sum_{i=1}^{2k}\sum_{j=i+1}^{2k+1}\cos\left(\frac{2(i-j)\pi}{2k+1}\right)=-k-\frac{1}{2}.$$
Expanding this double sum gives
\begin{align*}
\sum_{i=1}^{2k}\sum_{j=i+1}^{2k+1}\cos\left(\frac{2(i-j)\pi}{2k+1}\right) & = \cos\left(\frac{-2\pi}{2k+1}\right)+\cos\left(\frac{-4\pi}{2k+1}\right)+\ldots+\cos\left(\frac{-4k\pi}{2k+1}\right) \\
                    & + \cos\left(\frac{-2\pi}{2k+1}\right)+\cos\left(\frac{-4\pi}{2k+1}\right)+\ldots+ \cos\left(\frac{-(4k-2)\pi}{2k+1}\right) \\
                    &\vdots \\
                    & + \cos\left(\frac{-2\pi}{2k+1}\right)+\cos\left(\frac{-4\pi}{2k+1}\right) \\                           & +\cos\left(\frac{-2\pi}{2k+1}\right)
\end{align*}
which can be rearranged as:
$$\sum_{i=1}^{2k}\sum_{j=i+1}^{2k+1}\cos\left(\frac{2(i-j)\pi}{2k+1}\right) =2k\cos\left(\frac{2\pi}{2k+1}\right)+(2k-1)\cos\left(\frac{4\pi}{2k+1}\right)+\ldots+2\cos\left(\frac{(4k-2)\pi}{2k+1}\right)+\cos\left(\frac{4k\pi}{2k+1}\right).$$
Then, my question is the following: is it true that
$$ 2k\cos\left(\frac{2\pi}{2k+1}\right)+(2k-1)\cos\left(\frac{4\pi}{2k+1}\right)+\ldots+2\cos\left(\frac{(4k-2)\pi}{2k+1}\right)+\cos\left(\frac{4k\pi}{2k+1}\right)=-k-\frac{1}{2}?$$
I checked it for some values of $k$ and try induction for the general case, but I couldn't get very far. Also, as I mentioned, one can prove the result using Vieta's formula but it seems that one should be able to prove it this way as well.
 A: If you consider the points to be vectors, and lay them tail to head, you create a curve consisting of line segments of equal lengths with exterior angles of $\frac{2\pi}{n}$. A regular n-gon consists of line segments of equal lengths with exterior angles of $\frac{2\pi}{n}$. You can't have two different shapes made up of the same lengths and angles, so the curve must be a regular n-gon, which means that the last one joins up with the first one, which means their sum is $0$.
A: The answer by Golden_Ratio is totally fine, but here is a potentially more intuitive, less abstract way of seeing the answer using symmetry:
Say you have five points. If you rotate them all by a fifth of a full turn, the five points are in the same five positions, just shuffled in a different order.
The average of the five points therefore

*

*Has to be exactly the same, because the five points are the same

*Has to be rotated by a fifth of a full turn.

The only number that can work here (i.e. be rotated but stay exactly the same), is the number $0$. So this must be the asnwer.
A: The result should hold for any $n>1$ (not just odd $n$). First recall for $r\neq 1$,
$$\sum_{k=0}^{n-1} r^k=\frac{1-r^{n}}{1-r}.\quad (1)$$
Note that equally spaced points on the unit circle are essentially the roots of unity up to a rotation by angle $\phi$. And it is known that the sum of the $n$th roots of unity up to any rotation for $n>1$ is zero since
$$\sum_{k=0}^{n-1} e^{(2k\pi/n+\phi)i}
=e^{\phi i}\sum_{k=0}^{n-1} e^{2k\pi i/n}
=e^{\phi i}\frac{1-e^{2\pi i}}{1- e^{2\pi i/n}}=0,\\
$$
where the penultimate equality uses $(1)$, and the ultimate equality is by Euler's identity.
A: Symmetry permits no other average than the center of the circle.  Here's one way to make that somewhat more formal:

Choose an orthonormal coordinate system with its origin at the center of a circle with radius $r$, and the x axis directed such that $\theta_1 = 0$.
Observe that because of the even spacing, the remaining angles group in pairs of the form $\pm \frac{m2\pi}{n}$.  The members of each pair will have the opposite sines (and $\sin \theta_1 = 0$), therefore in the average, all the sine terms, which correspond proportionally to $y$ coordinates, cancel.  It follows that the average of all the points lies on the chosen x axis, a line containing the origin and $(r, \theta_1)_{polar}$.
Now observe that we can apply the same argument to determine that the average lies on each line containing the origin and any of the $(r, \theta_i)$.  These cannot all be the same line, because a line intersects a circle at at most two points, and there are at least three distinct $(r, \theta_i)$.  Choose any two distinct lines from the set. We know that the average lies on both, so it must be at their intersection.  We know that their intersection is unique, because the lines are distinct.  And we know that the origin, which is at the center of the circle, is on both lines, so it must be the intersection.

Note that that does not particularly depend on the number of points being odd.  With only minor changes, it can be generalized to any number of points greater than two.
