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Math Help - de moivre's thm

  1. #1
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    Sep 2007
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    de moivre's thm

    froom my textbook it writes:
    de moivre's thm
    (cosx + i sin x )^{\frac p q } = cos (\frac p q x + \frac {2k\pi} q ) + i sin (\frac p q x + \frac {2k\pi} q )
    for  k = 0, 1, 2, ... , q-1
    i don't understand why k cannot be greater than q-1. thanks
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  2. #2
    Member
    Joined
    May 2008
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    it can be, just if you do the solution you get will be exactly the same as one you have found before say for example if k=q
    we have
    <br />
\cos (\frac{p}{q}x+2\pi) +i \sin (\frac{p}{q}x+2\pi)

    this is identical to  \cos (\frac{p}{q}x) + i \sin (\frac{p}{q}x) which is the result with k=0
    (because sine and cosine are periodic functions with period 2pi)

    so k could be q,q+1...etc just you would just repeat all of the solutions
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