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Math Help - Solve with de moivres formula

  1. #1
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    Solve with de moivres formula

    a) (sqrt3 - i)^6 b) (2-2i)^5
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    Re: Solve with de moivres formula

    Quote Originally Posted by kalemale View Post
    a) (sqrt3 - i)^6 b) (2-2i)^5
    \sqrt3-i=2\exp\left(-\frac{i\pi}{6}\right) and 2-2i=\sqrt8(\exp\left(\frac{-i\pi}{4}\right)
    Last edited by Plato; July 17th 2012 at 03:25 AM.
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    Re: Solve with de moivres formula

    Quote Originally Posted by Plato View Post
    \sqrt3-i=2\exp\left(-\frac{\pi}{3}\right)
    OP, even experts make tiny mistakes sometimes.

    \sqrt3-i=2\exp\left(-\frac{\pi}{6} i \right)

    I hope i got that right.
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  4. #4
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    Re: Solve with de moivres formula

    sorry if I dont understand fully but why is it
    -pi/4 and -pi/6
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    Re: Solve with de moivres formula

    Quote Originally Posted by kalemale View Post
    sorry if I dont understand fully but why is it
    -pi/4 and -pi/6
    If you plot the complex number  \sqrt 3 - i on the complex plain and convert to polar coordinates you get R = 2 and \theta= \frac {- \pi}6. Stated another way:  \sqrt 3 - i = 2(\cos (\frac {-\pi} 6}) + i \sin  (\frac {-\pi} 6}) ) = 2 cis (\frac {-\pi} 6}). Now to raise it to the 6th power apply de Moivre's theorem:  (R cis  \theta) ^n = r^n cis(n\theta)
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    Re: Solve with de moivres formula

    so we get -pi / 6 becouse its negativ and the angle is 30 degress?
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    Re: Solve with de moivres formula

    Quote Originally Posted by kalemale View Post
    so we get -pi / 6 becouse its negativ and the angle is 30 degress?
    Yes - the angle is negative 30 degrees.
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