How Do You Calculate and Verify the Roots of a Complex Cubic Equation?

[rock.freak667]

Homework Statement

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

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

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

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

Homework Equations

complex number formulas

The Attempt at a Solution

z^3=-(4\sqrt{3})+4i

= z^3=8e^{\frac{5}{6}\pi i}z=2e^{\frac{5}{18}\pi i}

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

therefore the roots are

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

subbing the roots into what they want me to show(2e^{\frac{5}{18}\pi i})^{3k}+(2e^{\frac{17}{18}\pi i})^{3k}+(2e^{\frac{29}{18}\pi i})^{3k}

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

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

Now I am stuck.

 

[Rainbow Child]

...
(2e^{\frac{5}{18}\pi i})^{3k}+(2e^{\frac{17}{18}\pi i})^{3k}+(2e^{\frac{29}{18}\pi i})^{3k}

2^{3k}(e^{\frac{5k}{6}\pi i}+e^{\frac{17k}{6}\pi i}+e^{\frac{29k}{\mathbf{18}}\pi i}) ...mistake

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

Now I am stuck.

It should be

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}