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.

  • $\begingroup$ It could help to think in terms of the poles of a (lowpass prototype) Butterworth filter, which are exactly like that. As it turns out, for even orders you have an even number of symmetrically placed poles (both on x and y, if you consider the H(w)^2), and odd orders have an additional pole at -1 (and +1 for H(w)^2). So they would all cancel if summed. I suppose a picture would make it easier (only the left hand side). $\endgroup$ Feb 17, 2022 at 21:57
  • 3
    $\begingroup$ Using complex numbers you get a very simple and elegant proof. See Proof that sums of complex unit roots is zero $\endgroup$ Feb 17, 2022 at 23:39
  • 2
    $\begingroup$ For another geometric argument, that's equivalent to saying that the centroid ("average") of the vertices of a regular polygon is the center of its circumcircle. Consider that the centroid of a polygon does not change if you replace the endpoints of each side with the midpoint of the same side. Repeat this process, starting from the given regular polygon. At each step, you get a new, smaller regular polygon, with a circumcircle having the same center. In the limit, as the polygons become tinier and tinier, all vertices converge to the center of the circle, so that's the centroid. $\endgroup$
    – dxiv
    Feb 18, 2022 at 7:20
  • 1
    $\begingroup$ This is simply an instance of a finite geometric sequence summation in which the first-present and the first-absent terms of the underlying (infinite) geometric sequence have difference $0$. $\endgroup$ Feb 19, 2022 at 10:43

4 Answers 4


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

  1. Has to be exactly the same, because the five points are the same
  2. 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.

  • $\begingroup$ That's very nice indeed, the average of the rotatio is the rotation of the average... I had a bunch of geometric arguments in mind but this one is very clean. Thanks! $\endgroup$
    – hamath
    Feb 17, 2022 at 14:36
  • 19
    $\begingroup$ Pedant Alert: 1 is an odd number (of points on the circle) whose average position is not at the origin, but on the circle wherever it's at. $\endgroup$
    – user121330
    Feb 17, 2022 at 21:18
  • 2
    $\begingroup$ In other words, $\bar x = e^{(2\pi i)/n} \bar x$, so $(1- e^{(2\pi i)/n} )\bar x =0$. If $n=1$, this just gives $0 = 0$, but if $n>1$, it follows that $\bar x =0$. $\endgroup$ Feb 19, 2022 at 7:54
  • $\begingroup$ @user121330: It's definitely valuable to point out that in this argument, having a non-identity rotation is key! It's can also be reasonable to leave a step like that implicit. This answer is written at the same level of formality as the question, which also says the average should be zero even though that doesn't happen for a single point. Whether the inconsistency is worth flagging depends on whether it's a misconception or an intentional informality—and also whether it's likely to mislead bystanders. $\endgroup$
    – Vectornaut
    Feb 20, 2022 at 0:08

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.

  • $\begingroup$ Interestingly, the formula for the geometric sum is usually seen by multiplying by $1-r$, which connects this to the answer by Simon. $\endgroup$
    – Carsten S
    Feb 18, 2022 at 15:11
  • 3
    $\begingroup$ "The result should hold for any n>1 (not just odd n)." It's not that the proof is harder for odd; rather, the proof is much more trivial for even; each point can be paired with an opposite that cancels it out. That leaves the odd case as needing a further proof. $\endgroup$ Feb 19, 2022 at 7:34
  • $\begingroup$ @Acccumulation: Yes — but having given the proof for the odd case, one sees it works for the general case, so there’s no need to do a case split in the first place. $\endgroup$ Feb 19, 2022 at 20:38

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.

  • 1
    $\begingroup$ +1. Perhaps more informally and shorter: draw a line through one of the points and the centre, and see that the average must lie on this line by symmetry of the pairs of other points across the line; since it lies on all such lines it must be where such lines intersect at the centre, at least when $n>2$ when there is more than one line. By inspection, the statement is also true when $n=2$ (a different symmetry) but not true when $n=1$ $\endgroup$
    – Henry
    Feb 18, 2022 at 10:43
  • $\begingroup$ That's very nice! I actually had the argument for the sines in mind, but I didn't know how proceed for the average on the x axis. Thanks! $\endgroup$
    – hamath
    Feb 18, 2022 at 12:39

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 $0$.

  • $\begingroup$ The eleventh word ought to be "them." $\endgroup$
    – phoog
    Feb 20, 2022 at 11:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .