# Roots of unity

• Apr 21st 2008, 03:45 AM
biplab
Roots of unity
I have n roots of unity a1,a2,.....an...the wat is the value of the following??
(1-a1)(1-a2)....(1-an)...
• Apr 21st 2008, 04:12 AM
mr fantastic
Quote:

Originally Posted by biplab
I have n roots of unity a1,a2,.....an...the wat is the value of the following??
(1-a1)(1-a2)....(1-an)...

One of the roots is equal to 1, that is, $a_k = 1$ for a value of k between 1 and n inclusive. Therefore ......
• Apr 21st 2008, 05:18 PM
ThePerfectHacker
Quote:

Originally Posted by biplab
I have n roots of unity a1,a2,.....an...the wat is the value of the following??
(1-a1)(1-a2)....(1-an)...

I assume you want to find $(1-a_1)...(1-a_{n-1})$ instead where $a_i\not = 1$ are the n roots of unity. Because otherwise like Mr.Fantastic said the problem is trivial.

Note, $x^n - 1 = \prod_{k=0}^{n-1} (x - \zeta^k)$ where $\zeta = e^{2\pi i/n}$. This means, $1+x+...+x^{n-1} = \prod_{k=1}^{n-1} ( x - \zeta^k)$. Let $x=1$ and so $n = (1- a_1)...(1-a_{n-1})$.
• Apr 21st 2008, 09:49 PM
mr fantastic
Quote:

Originally Posted by ThePerfectHacker
I assume you want to find $(1-a_1)...(1-a_{n-1})$ instead where $a_i\not = 1$ are the n roots of unity. Because otherwise like Mr.Fantastic said the problem is trivial.

Note, $x^n - 1 = \prod_{k=0}^{n-1} (x - \zeta^k)$ where $\zeta = e^{2\pi i/n}$. This means, $1+x+...+x^{n-1} = \prod_{k=1}^{n-1} ( x - \zeta^k)$. Let $x=1$ and so $n = (1- a_1)...(1-a_{n-1})$.

A small correction/clarification/expansion:

$x^n - 1 = \prod_{k=0}^{n} (x - \zeta^k)$ where $\zeta = e^{2\pi i/n}$

$= (x - 1) \prod_{k=0}^{n-1} (x - \zeta^k)$.

But $x^n - 1 = (x - 1)(1+x+...+x^{n-1})$.

Therefore $(x - 1)(1+x+...+x^{n-1}) = (x - 1) \prod_{k=0}^{n-1} (x - \zeta^k)$

$\Rightarrow 1+x+...+x^{n-1} = \prod_{k=1}^{n-1} ( x - \zeta^k)$

etc.

By the way, $\prod$ is the product symbol (Multiplication - Wikipedia, the free encyclopedia), the multiplicative analogue of the summation symbol $\Sigma$ (Summation - Wikipedia, the free encyclopedia)