# Products of the unit circle

• Apr 6th 2007, 10:49 AM
Products of the unit circle
Here is the problem:
Let P sub 0, ....., P sub n-1 be n equally spaced points on the unit circle. Compute the product of the distances from P sub 0 (naught) to all the remaining points.
• Apr 6th 2007, 11:31 AM
ThePerfectHacker
Quote:

Here is the problem:
Let P sub 0, ....., P sub n-1 be n equally spaced points on the unit circle. Compute the product of the distances from P sub 0 (naught) to all the remaining points.

I do not understand. What distances and what products.

First we know that,
P_k = exp((2pi/n)*ki)

Now what?
• Apr 6th 2007, 11:42 AM
Plato
• Apr 6th 2007, 01:35 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
This is my understanding.

In that case the product is equal to "n" itself.
• Apr 6th 2007, 02:43 PM
Plato
Quote:

Originally Posted by ThePerfectHacker
In that case the product is equal to "n" itself.

Correct! What is the proof?
• Apr 8th 2007, 07:51 AM
ThePerfectHacker
Quote:

Originally Posted by Plato
Correct! What is the proof?

Not an easy question to answer.
But we can begin by simplifing.
• Apr 8th 2007, 08:53 AM
ThePerfectHacker
I am going in unit circles (get the pun) with this problem.

But I managed to make this problem more complicated than it should be. Instead of working with cyclotonomic polynomials, as I should have. I reduced the problem to trigonometry which makes it more difficult.

I am sure you know what I am about to say but let me just post it.

A cyclotonomic polynomial is,
\Phi_n(z) = 1+z+z^2+...+z^n

We can factor them as,
(z-z_1)(z-z_2)+...+(z-_n)
Where,
z_1,z_2,...,z_n
Are points on the regular (n+1)-gon inscribed in the unit circle in the complex plane (except unity).

Basically, I am saying this is the cyclotonomic polynomial of degree n evaluated at 1.