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Homework Help: Roots of complex numbers

  1. Jan 27, 2008 #1


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    Homework Helper

    1. The problem statement, all variables and given/known data
    Find the roots of the equation

    giving your answers in the form [itex]re^{i\theta}[/itex], where r>0 and [itex]0\leq \theta<2\pi[/itex]

    Denoting these roots by [itex]z_1,z_2,z_3[/itex], show that, for every positive integer k.

    [tex]z_1^{3k}+z_2^{3k}+z_3^{3k}=3(2^{3k}e^{\frac{5}{6}k\pi i})[/tex]

    2. Relevant equations

    complex number formulas

    3. The attempt at a solution


    [tex]= z^3=8e^{\frac{5}{6}\pi i}[/tex]

    [tex]z=2e^{\frac{5}{18}\pi i}[/tex]

    [tex]z=2e^{(\frac{5}{18}\pi + \frac{2k}{3})i}[/tex] k=0,1,2

    therefore the roots are

    [tex]z=2e^{\frac{5}{18}\pi i},2e^{\frac{17}{18}\pi i},2e^{\frac{29}{18}\pi i}[/tex]

    subbing the roots into what they want me to show

    [tex](2e^{\frac{5}{18}\pi i})^{3k}+(2e^{\frac{17}{18}\pi i})^{3k}+(2e^{\frac{29}{18}\pi i})^{3k}[/tex]

    [tex]2^{3k}(e^{\frac{5k}{6}\pi i}+e^{\frac{17k}{6}\pi i}+e^{\frac{29k}{18}\pi i})[/tex]

    [tex]2^{3k}e^{\frac{5}{6}\pi i}(1+e^{2k}+e^{4k})[/tex]

    Now I am stuck.
  2. jcsd
  3. Jan 27, 2008 #2
    It should be

    [tex]e^{\frac{17k}{6}\pi i}=e^{\frac{5k}{6}\pi i}\,e^{\frac{12k}{6}\pi i}=e^{\frac{5k}{6}\pi i}\,e^{2\,k\,\pi i}=e^{\frac{5k}{6}\pi i}[/tex]
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