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 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 , with integer k=0,1,…,n-1. Or, if we define , we have roots 1, ζ, ζ^{2},…, ζ^{n-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 .

Thus .

However, we know that

Thus the polynomial

Now, the distance between the root ζ^{k} and 1 is

Thus the product of the distances from 1 to each of the other roots is:

But as , we see

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 . 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: . 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 , and so our product is .

*n*=4: This is a square with sides of length , and thus diagonals of length 2, and so the product is .

*n*=6: This is a hexagon with sides of length 1. It has six diagonals of length , and three of length 2. Thus the product is .

and so on.

Now, let us examine the roots of unity again. We said , and so . Now, using Euler’s formula, this is

;

and thus

as .

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 for all *x* to get that , , et cetera, to show that for odd *n*:

.

Examples:

and so on.

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This entry was posted on February 25, 2008 at 6:54 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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March 14, 2008 at 7:31 am |

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

November 1, 2010 at 12:21 am |

[…] 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 […]

April 13, 2011 at 3:20 pm |

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