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Math Help - Roots of unity

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    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)...
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    Quote Originally Posted by biplab View Post
    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 ......
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    Quote Originally Posted by biplab View Post
    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}).
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    Quote Originally Posted by ThePerfectHacker View Post
    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)
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