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Math Help - De Moivre Theorem Help

  1. #1
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    De Moivre Theorem Help

    I have to evaluate,
    (1-\sqrt3i)^5

    so far I have that I found,
    r(\cos\theta + i\sin\theta)
    where r = 2

    and that tan (theta) = -\sqrt3

    but where do i go now?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by driverfan2008 View Post
    I have to evaluate,
    (1-\sqrt3i)^5

    so far I have that I found,
    r(\cos\theta + i\sin\theta)
    where r = 2

    and that tan (theta) = -\sqrt3

    but where do i go now?
    Solving for \theta, you should have \theta=-\frac{\pi}{3}

    Therefore, the complex number in polar form is the same as 2\left[\cos\left(-\tfrac{\pi}{3}\right)+i\sin\left(-\tfrac{\pi}{3}\right)\right]=2e^{i\left(-\frac{\pi}{3}\right)}

    Thus, \left(1-\sqrt{3}i\right)^5=\left(2e^{i\left(-\frac{\pi}{3}\right)}\right)^5=32e^{i\left(-\frac{5\pi}{3}\right)}=32e^{i\frac{\pi}{3}} =32\left[\cos\left(\tfrac{\pi}{3}\right)+i\sin\left(\tfrac{  \pi}{3}\right)\right]=\boxed{16+16\sqrt{3}i}
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  3. #3
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    How did you go from,

    =32\left[\cos\left(\tfrac{\pi}{3}\right)+i\sin\left(\tfrac{  \pi}{3}\right)\right]

    to

    =\boxed{16+16\sqrt{3}i}
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  4. #4
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    Quote Originally Posted by driverfan2008 View Post
    How did you go from,

    =32\left[\cos\left(\tfrac{\pi}{3}\right)+i\sin\left(\tfrac{  \pi}{3}\right)\right]

    to

    =\boxed{16+16\sqrt{3}i}
    What's \cos{\frac{\pi}{3}} equal?

    What's \sin{\frac{\pi}{3}} equal?
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