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Math Help - Roots of unity Question

  1. #1
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    Roots of unity Question

    Hello, can I get some help with the following?

    Show that if w is an nth root of unity (w != 1 and n > 1) then
    w + w^2 + ...+ w^n = 0

    Thanks!
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  2. #2
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    Quote Originally Posted by noobonastick View Post
    Hello, can I get some help with the following?

    Show that if w is an nth root of unity (w != 1 and n > 1) then
    w + w^2 + ...+ w^n = 0

    Thanks!
    If w = \sqrt[n]{1 + 0i}

    Then w^n = 1 + 0i

    w^n = e^{2\pi i}.


    This means by DeMoivre's Theorem, the focus root is

    w = e^{\frac{2\pi i}{n}}.

    But there must be exactly n roots, all spaced evenly about a circle.

    So that means any w can be written as

    w = e^{\frac{2k\pi i}{n}}, where k is a positive integer no greater than n.


    That means

    w^2 = e^{\frac{4k\pi i}{n}}

    w^3 = e^{\frac{6k\pi i}{n}}

    \vdots

    w^n = e^{2k\pi i}

    But any integer multiple of 2\pi gives the same result as 2\pi.

    So that means w^n = e^{2\pi i}.



    Now let's go back to the original problem...

    We have w + w^2 + w^3 + \dots + w^n.

    This is a geometric series with a = w and r = w.


    So the sum is

    \frac{a(1 - r^n)}{1 - r}

     = \frac{w(1 - w^n)}{1 - w}

     = \frac{e^{\frac{2k\pi i}{n}}\left(1 - e^{2\pi i}\right)}{1 - e^{\frac{2k\pi i}{n}}}.


    But e^{2\pi i} = 1.

    So that means the sum is

    \frac{e^{\frac{2k\pi i}{n}}\left(1 - 1\right)}{1 -  e^{\frac{2k\pi i}{n}}}

     = \frac{0e^{\frac{2k\pi i}{n}}}{1 -  e^{\frac{2k\pi i}{n}}}

     = 0.


    Q.E.D.
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  3. #3
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    Quote Originally Posted by noobonastick View Post
    Hello, can I get some help with the following?

    Show that if w is an nth root of unity (w != 1 and n > 1) then
    w + w^2 + ...+ w^n = 0

    Thanks!
    I w is an nth root of unity then w^n= 1, of course, so that equation becomes 1+ w+ w^2+ ...+ w^{n-1}= 0.

    Also, since w^n= 1, w^n- 1= (w- 1)(1+ w+ w^2+ ...+ w^{n-1})= 0.
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