Monday Math 9: Points on the unit circle

Place n equally-spaced points on the unit circle, n≥3. Draw all of the line segments connecting any two of these points (the regular n-gon and it’s diagonals). What is the product of the lengths of all \frac{n(n-1)}{2} of these line segments?

By using complex numbers to represent points in the plane, we can not only solve this problem in the case of the general n≥3, but we can derive some interesting trigonometric identities as well.

In the complex plane, there are n nth-roots of unity, which can be given as z=e^{i\frac{2\pi{k}}{n}}, with integer k=0,1,…,n-1. Or, if we define \zeta=e^{i\frac{2\pi}{n}}, we have roots 1, ζ, ζ2,…, ζn-1.
The absolute value of all of these roots are unity, and we see that if we plot them on the complex plane, they are n points equally spaced on the unit circle.

Note that these are the roots of the polynomial z^n-1.
Thus z^n-1=(z-1)(z-\zeta)(z-\zeta^2)\ldots(z-\zeta^{n-1}).
However, we know that z^n-1=(z-1)(1+z+z^2+\ldots+z^{n-1})
Thus the polynomial

Now, the distance between the root ζk and 1 is \left|1-\zeta^k\right|
Thus the product of the distances from 1 to each of the other roots is:
But as \Phi(z)=1+z+z^2+\ldots+z^{n-1}, we see \Phi(1)=1+1+1^2+\ldots+1^{n-1}=n
So the product of the distances from one point to all the n-1 others is simply n.
If we perform this product for each of the n points and multiply those, we obtain n^n. However, in that product, each length appears twice (once for each endpoint being the ‘fixed’ point). Thus the desired product is the square root of that: n^{\frac{n}{2}}. We can confirm this for a few basic examples:
n=3: This is just an equilateral triangle in the unit circle. Each side is of length \sqrt{3}, and so our product is \sqrt{3}\cdot\sqrt{3}\cdot\sqrt{3}=3^{\frac{3}{2}}.
n=4: This is a square with sides of length \sqrt{2}, and thus diagonals of length 2, and so the product is (\sqrt{2})^4\cdot2^2=16=4^2=4^{\frac{4}{2}}.
n=6: This is a hexagon with sides of length 1. It has six diagonals of length \sqrt{3}, and three of length 2. Thus the product is 1^6\cdot(\sqrt{3})^6\cdot2^3=216=6^3=6^{\frac{6}{2}}.
and so on.

Now, let us examine the roots of unity again. We said \zeta=e^{i\frac{2\pi}{n}}, and so \zeta^k=e^{i\frac{2\pi{k}}{n}}. Now, using Euler’s formula, this is
and thus
\begin{array}{rcl}\left|\zeta^k-1\right|&=&\left|\cos\left(\frac{2\pi{k}}{n}\right)+i\sin\left(\frac{2\pi{k}}{n}\right)-1\right|\\  &=&\sqrt{\left(\cos\left(\frac{2\pi{k}}{n}\right)-1\right)^2+\left(\sin\left(\frac{2\pi{k}}{n}\right)\right)^2}\\  &=&\sqrt{\cos^2\left(\frac{2\pi{k}}{n}\right)-2\cos\left(\frac{2\pi{k}}{n}\right)+1+\sin^2\left(\frac{2\pi{k}}{n}\right)}\\  &=&\sqrt{2-2\cos\left(\frac{2\pi{k}}{n}\right)}\\  &=&\sqrt{2\left(2\sin^2\left(\frac{\pi{k}}{n}\right)\right)}\\  &=&2\sin\left(\frac{\pi{k}}{n}\right)\end{array}
as 0\le{k}\le{n-1}.
Thus, we see
and we know from our previous work that this that this product equals n. Therefore,

For example, this says:

Note also, we can combine this for odd n with the fact that \sin(\pi-x)=\sin{x} for all x to get that \sin(\frac{\pi(n-1)}{n})=\sin(\frac{\pi}{n}), \sin(\frac{\pi(n-2)}{n})=\sin(\frac{2\pi}{n}), et cetera, to show that for odd n:
and so on.


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3 Responses to “Monday Math 9: Points on the unit circle”

  1. twistedone151 Says:

    There, fixed it. For some reasons, all the backslashes in the TeX markup were removed.

  2. Monday Math 141 « Twisted One 151's Weblog Says:

    […] Steinbach, who outlined a proof using the n-gon formed by the complex n-roots of unity (compare this problem, for example). Now, we can generalize the formula first by exchanging k and m to see that for […]

  3. Math Problem of the Day » Problem of the Day #24: Product of areas of triangles in regular n-gon Says:

    […] the distances from one vertex to all the other vertices is the number of sides of the polygon (cf. this result), we have $prod_{r=1}^{n-1} A_n A_{r} = nR^{n-1}$, so $P^2 = (n R^{n-1})^{n(n-2)}$, so $P = […]

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