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Math Help - De moivres theorem.

  1. #1
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    De moivres theorem.

    Use DMT to solve the following equations in polar form.

     z^3=-1-i
    z^3=-4+4\sqrt{3i}
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    What have you tried so far?. These problems are just routine knowing the corresponding theory.
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  3. #3
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    I'll show you how to do one, then you try to do the other...

    \displaystyle |-1-i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}.

    Note that this complex number is in the third quadrant. The angle in the first quadrant is \displaystyle \arctan{\frac{1}{1}} = \frac{\pi}{4}.
    So the angle in the third quadrant, \displaystyle \arg{(-1 - i)} = \pi + \frac{\pi}{4} = \frac{5\pi}{4}.


    \displaystyle z^3 = \sqrt{2}\,\textrm{cis}\,{\left(\frac{5\pi}{4}\righ  t)}

    \displaystyle z = (\sqrt{2})^{\frac{1}{3}}\,\textrm{cis}\,{\left(\fr  ac{1}{3}\cdot \frac{5\pi}{4}\right)}

    \displaystyle = \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{5\pi}{12}\right)}.


    There are always three cube roots, and they are of the same magnitude and evenly spaced about a circle. So they differ by an angle of \displaystyle \frac{2\pi}{3}.

    So the other solutions are \displaystyle \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{5\pi}{12} + \frac{2\pi}{3}\right)} = \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{13\pi}{12}\right)}

    and \displaystyle \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{13\pi}{12} + \frac{2\pi}{3}\right)} = \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{21\pi}{12}\right)}.


    But since \displaystyle -\pi < \arg{z} \leq \pi, we rewrite this as

    \displaystyle \sqrt[6]{2}\,\textrm{cis}\,{\left(-\frac{11\pi}{12}\right)} and \displaystyle \sqrt[6]{2}\,{\left(-\frac{3\pi}{12}\right)}.
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