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Roots of complex numbers

  • #1
rock.freak667
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Homework Statement


Find the roots of the equation
[tex]z^3=-(4\sqrt{3})+4i[/tex]

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]

Homework Equations



complex number formulas

The Attempt at a Solution



[tex]z^3=-(4\sqrt{3})+4i[/tex]

[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.
 

Answers and Replies

  • #2
...
[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}{\mathbf{18}}\pi i})[/tex] ....mistake

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

Now I am stuck.
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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