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Math Help - DeMoivre's theorem

  1. #1
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    DeMoivre's theorem

    "use DeMoivre's theorem to find the indicated power of the complex number. Write the answer in rectangular form."

    (squareroot{3(cos 5pie/6 + isin 5pie/6))^4

    Am I doing this correctly?

    (sqrt3)^4(cos(4*5pie/6)+isin(4*5pie/6)
    9(cos 20pie/6)+isin(20pie/6)
    9(-.5)+(-.866i)
    =-4.5-7.8i

    I'm worried I might have done this wrong due to the step in reducing 20pie/6 I couldn't find anything to reduce it by so just left it in that form.
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  2. #2
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    Quote Originally Posted by ConfusedMath View Post
    "use DeMoivre's theorem to find the indicated power of the complex number. Write the answer in rectangular form."

    (squareroot{3(cos 5pie/6 + isin 5pie/6))^4

    Am I doing this correctly?

    (sqrt3)^4(cos(4*5pie/6)+isin(4*5pie/6)
    9(cos 20pie/6)+isin(20pie/6)
    9(-.5)+(-.866i)
    =-4.5-7.8i

    I'm worried I might have done this wrong due to the step in reducing 20pie/6 I couldn't find anything to reduce it by so just left it in that form.
    \left[\sqrt{3}\left(\cos{\frac{5\pi}{6}} + i\sin{\frac{5\pi}{6}}\right)\right]^4 = (\sqrt{3})^4 \left(\cos{\frac{4\cdot 5\pi}{6}} + i\sin{\frac{4\cdot 5\pi}{6}}\right)

     = 9\left(\cos{\frac{10\pi}{3}} + i\sin{\frac{10\pi}{3}}\right)

     = 9\left(\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}}\right)

     = 9\left(-\cos{\frac{\pi}{3}} - i\sin{\frac{\pi}{3}}\right)

     = 9\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)

     = -\frac{9}{2} - \frac{9\sqrt{3}}{2}i.
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