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N-th root of unity

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that [itex]z=e^{\frac{2k\pi i}{n}},n\in\mathbb{N},k\in\mathbb{Z}, 0\leq k\leq n-1 [/itex] is an n-th root of unity.

    3. The attempt at a solution

    So I know I have to come to the conclusion that [itex]z^{n}=1[/itex]. I'm thinking of using the property [itex]e^{i\theta}=cos\theta+isin\theta[/itex], but when I try to break it up like that I get something strange like [itex]e^{\frac{2k\pi i}{n}}=cos(\frac{2k\pi}{n})+i( \frac{2k\pi}{n})[/itex] and I have to get it into a form like [itex]cos^{2}(\theta)+sin^{2}(\theta) [/itex]. Any ideas on how to do that/am I on the right track?
     
  2. jcsd
  3. Jan 19, 2012 #2

    SammyS

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    Look at [itex]\left(e^{\displaystyle \left(\frac{2k\pi i}{n}\right)}\right)^n\,.[/itex]
     
  4. Jan 19, 2012 #3
    Oh, so I get [itex]\left(e^{{\displaystyle \left(\frac{2k\pi i}{n}\right)}}\right)^{n}=e^{2k\pi i}=cos(2k\pi)+isin(2k\pi)... [/itex] which looks better. But how do I go from there?
     
  5. Jan 19, 2012 #4

    SammyS

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    If k is an integer, that is equal to 1 .
     
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