How do you find the three cube roots of -2-2i \sqrt{3}?

Oct 24, 2014

Let $z = r {e}^{i \theta}$ be the three roots.

${z}^{3} = - 2 - 2 i \sqrt{3} = 4 \left(- \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)$

by rewriting in exponential form,

$R i g h t a r r o w {r}^{3} {e}^{i \left(3 \theta\right)} = 4 {e}^{i \left(\frac{4 \pi}{3} + 2 n \pi\right)}$,

where $n$ is any integer,

$R i g h t a r r o w \left\{\begin{matrix}{r}^{3} = 4 R i g h t a r r o w r = \sqrt[3]{4} \\ 3 \theta = \frac{4 \pi}{3} + 2 n \pi R i g h t a r r o w \theta = \frac{4 \pi}{9} + \frac{2 n \pi}{3}\end{matrix}\right.$

Hence, the three roots are

$z = \setminus \left\{\sqrt[3]{4} {e}^{i \frac{4 \pi}{9}} , \sqrt[3]{4} {e}^{i \frac{10 \pi}{9}} , \sqrt[3]{4} {e}^{i \frac{16 \pi}{9}}\right\}$.

I hope that this was helpful.