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Thread: De Moivre's Theorem

  1. #1
    Junior Member masoug's Avatar
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    Question De Moivre's Theorem

    Hi,
    I was reading a textbook when I came upon their discussion on De Moivre's Theorem. The book talked about the difference between $\displaystyle (-8)^{\frac{1}{3}}$ and $\displaystyle (-8)^{\frac{2}{6}}$, and they said something about how De Moivre's Theorem let us conclude that there are 3 cube roots of a number and 6 sixth roots. Can anyone decipher what this means?

    Thanks!

    -Masoug
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  2. #2
    MHF Contributor
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    $\displaystyle \displaystyle w^n=z$

    It is obvious that $\displaystyle w=-2$ is a solution.

    $\displaystyle r=\sqrt{(-8)^2}=8$

    The angle of -8 is $\displaystyle \pi$ since it lies on the negative x axis.

    $\displaystyle \displaystyle w_k=r^{1/n}\left(cos\left(\frac{\theta +2\pi k}{n}\right)+\mathbf{i}sin\left(\frac{\theta +2\pi k}{n}\right)\right)$

    Where k goes from 0 to 1 less than the n so k = 0, 1, 2 and n = 3.

    Principle root: $\displaystyle \displaystyle w_0=8^{1/3}\left(cos\left(\frac{\pi+2\pi*0}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*0}{3}\right)\right)=\cdots$

    $\displaystyle \displaystyle w_1=8^{1/3}\left(cos\left(\frac{\pi+2\pi*1}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*1}{3}\right)\right)=\cdots$

    $\displaystyle \displaystyle w_2=8^{1/3}\left(cos\left(\frac{\pi+2\pi*2}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*2}{3}\right)\right)=\cdots$
    Last edited by dwsmith; Dec 18th 2010 at 12:57 PM. Reason: Changed w to z
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  3. #3
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    -8 has one real cube root (-2 as dwsmith says) and two complex roots. -8 has six complex (non-real) roots but if you square those, you will find that there are three sets of pairs that have the same square-giving you again the 3 cube roots of -8. For example, two of the sixth roots of -8 are $\displaystyle i\sqrt{2}$ and $\displaystyle -i\sqrt{2}$ both of which have square equal to -2.
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