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Thread: 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 $\displaystyle w = \sqrt[n]{1 + 0i}$

    Then $\displaystyle w^n = 1 + 0i$

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


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

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

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

    So that means any $\displaystyle w$ can be written as

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


    That means

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

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

    $\displaystyle \vdots$

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

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

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



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

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

    This is a geometric series with $\displaystyle a = w$ and $\displaystyle r = w$.


    So the sum is

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

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

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


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

    So that means the sum is

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

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

    $\displaystyle = 0$.


    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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