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Math Help - Simple Polar number question...i think...

  1. #1
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    Simple Polar number question...i think...

    Its quite simple but im not sure of it basically:
    Find the argument of
    1 + itan(\frac{2pi}{3})

    So tan is
    \frac{sin}{cos}
    So i change tan to this and multiply through and get
    cos(\frac{2pi}{3}) + isin(\frac{2pi}{3})

    So i get the answer, the argument is 2pi / 3...seems pretty obvious...except its not...its -1pi / 3

    :S, why is this?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by AshleyT View Post
    Its quite simple but im not sure of it basically:
    Find the argument of
    1 + itan(\frac{2pi}{3})

    So tan is
    \frac{sin}{cos}
    So i change tan to this and multiply through and get
    cos(\frac{2pi}{3}) + isin(\frac{2pi}{3})

    So i get the answer, the argument is 2pi / 3...seems pretty obvious...except its not...its -1pi / 3

    :S, why is this?
    \tan \dfrac{2\pi}{3} = -\sqrt{3} so that you have 1-i\sqrt{3}..

    and yeah, \tan \theta = \dfrac{-\sqrt{3}}{1}.. and what's arg?
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  3. #3
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    Quote Originally Posted by kalagota View Post
    \tan \dfrac{2\pi}{3} = -\sqrt{3} so that you have 1-i\sqrt{3}..

    and yeah, \tan \theta = \dfrac{-\sqrt{3}}{1}.. and what's arg?
    Thank's for your reply
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