I have n roots of unity a1,a2,.....an...the wat is the value of the following??

(1-a1)(1-a2)....(1-an)...

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- Apr 21st 2008, 03:45 AMbiplabRoots 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 AMmr fantastic
- Apr 21st 2008, 05:18 PMThePerfectHacker
I assume you want to find $\displaystyle (1-a_1)...(1-a_{n-1})$ instead where $\displaystyle a_i\not = 1$ are the n roots of unity. Because otherwise like Mr.Fantastic said the problem is trivial.

Note, $\displaystyle x^n - 1 = \prod_{k=0}^{n-1} (x - \zeta^k)$ where $\displaystyle \zeta = e^{2\pi i/n}$. This means, $\displaystyle 1+x+...+x^{n-1} = \prod_{k=1}^{n-1} ( x - \zeta^k)$. Let $\displaystyle x=1$ and so $\displaystyle n = (1- a_1)...(1-a_{n-1})$. - Apr 21st 2008, 09:49 PMmr fantastic
A small correction/clarification/expansion:

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

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

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

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

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

etc.

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