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

$$e^{\frac{17k}{6}\pi i}=e^{\frac{5k}{6}\pi i}\,e^{\frac{12k}{6}\pi i}=e^{\frac{5k}{6}\pi i}\,e^{2\,k\,\pi i}=e^{\frac{5k}{6}\pi i}$$