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Thread: demostration polynomial

  1. #1
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    demostration polynomial

    Hl! I have problems with demostration

    Let \zeta=e^{2\pi i/n} be a primitive nth root unity


    For numbers a and b, prove that
    a^n-b^n= (a-b) (a-\zeta b)(a- {\zeta}^2b) \cdot{} \cdot{} \cdot{} (a-\zeta^{n-1}b)


    and if n is odd , that
    a^n+b^n= (a+b) (a+\zeta b)(a+ {\zeta}^2b) \cdot{} \cdot{} \cdot{} (a+\zeta^{n-1}b)


    Hint: Set x=a/b if b\neq{0}


    I do not know how to start or how to use the hint
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  2. #2
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    Re: demostration polynomial

    $a^n - b^n = b^n \left(\left(\dfrac a b\right)^n - 1\right)$

    $x = \dfrac a b,~b \neq 0$

    $a^n - b^n = b^n (x^n - 1)$

    Consider the polynomial $x^n-1$

    It's roots are precisely the $n$ roots of unity and further if $\zeta = e^{2\pi i \frac \theta n}$ then these roots are given by $\zeta^k,~k \in 0,~1,~2,\dots n-1$

    thus we can write

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

    $a^n - b^n = b^n \displaystyle \prod_{k=1}^{n-1} \left(x-\zeta^k\right)$

    and you should be able to take it from here.
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  3. #3
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    Re: demostration polynomial

    There's a little more to it than that. If \zeta = \frac{2\pi i}{n} where i divides n, it is not a generator for the full set of roots.
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  4. #4
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    Re: demostration polynomial

    Quote Originally Posted by Archie View Post
    There's a little more to it than that. If \zeta = \frac{2\pi i}{n} where i divides n, it is not a generator for the full set of roots.
    I should have explicitly noted this but I believe the fact that $\zeta$ is stated to be a primitive root overcomes this.
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  5. #5
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    Re: demostration polynomial

    e^{2\pi  i/n} is a primitive n-th root of unity

    Here i=\sqrt{-1}
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  6. #6
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    Re: demostration polynomial

    thanks to all
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