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Math Help - complex number equation

  1. #1
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    complex number equation

    I'm trying to get the 4 solutions for this:
     w^4 = -8 +(8*3^\frac{1}{2})i

    In cis form and after applying de Moivre's theorem I get:

     2cis( \frac{-\pi}{12} +\frac{1}{2}k\pi)

    Where k is varying integer.

    I've checked several times and I can't see where I'm wrong.
    Help appreciated!
    Last edited by elieh; September 24th 2011 at 10:55 AM.
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  2. #2
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    Re: complex number equation

    Quote Originally Posted by elieh View Post
    I'm trying to get the 4 solutions for this:
     w^4 = -8 +8*3^\frac{1}{2}

    In cis form and after applying de Moivre's theorem I get:

     2cis( \frac{-\pi}{12} +\frac{1}{2}k\pi)

    Where k is varying integer.

    I've checked several times and I can't see where I'm wrong.
    Help appreciated!
    Have you written that correctly?
    OR should it be  w^4 = -8 +8\sqrt3\color{red}\mathbf{i}
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  3. #3
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    Re: complex number equation

    Sorry! my bad,
    I edited it
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  4. #4
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    Re: complex number equation

    Quote Originally Posted by elieh View Post
    I'm trying to get the 4 solutions for this:
     w^4 = -8 +(8*3^\frac{1}{2})i
     2cis( \frac{-\pi}{12} +\frac{1}{2}k\pi)
    You have the argument wrong.
    The only thing to add is k=0,1,2,3
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  5. #5
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    Re: complex number equation

    But I don't get whole numbers to put in my answer which should be in the form a+bi
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  6. #6
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    Re: complex number equation

    Quote Originally Posted by elieh View Post
    But I don't get whole numbers to put in my answer which should be in the form a+bi
    \text{Arg}\left( { - 8 + 8\sqrt 3 i} \right) = \pi  + \arctan \left( {\frac{{8\sqrt 3 }}{{ - 8}}} \right) = \frac{{2\pi }}{3}

    Now \frac{2\pi}{3}\frac{1}{4}=\frac{\pi}{6}
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  7. #7
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    Re: complex number equation

    That works, Thanks.
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