Thread: 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.

Originally Posted by chadlyter

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?

This is my understanding.

Originally Posted by Plato

This is my understanding.

In that case the product is equal to "n" itself.

Originally Posted by ThePerfectHacker

In that case the product is equal to "n" itself.

Correct! What is the proof?

Originally Posted by Plato

Correct! What is the proof?

Not an easy question to answer.
But we can begin by simplifing.

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.