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.

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.
