How do you find the cube root of -4sqrt2(1-i)?

May 26, 2017

Principal root:

$\sqrt[3]{- 4 \sqrt{2} \left(1 - i\right)} = \sqrt{2} + \sqrt{2} i$

There are two other roots.

Explanation:

By de Moivre's formula:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

If $n$ is a positive integer, $r > 0$ and $\theta \in \left(- \pi , \pi\right]$ then (by convention) the principal $n$th root of:

$r \left(\cos \theta + i \sin \theta\right)$

is:

$\sqrt[n]{r} \left(\cos \left(\frac{\theta}{n}\right) + i \sin \left(\frac{\theta}{n}\right)\right)$

There are $n - 1$ other non-principal $n$th roots.

Note that:

$- 4 \sqrt{2} \left(1 - i\right) = 8 \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\right)$

$\textcolor{w h i t e}{- 4 \sqrt{2} \left(1 - i\right)} = {2}^{3} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

Hence:

$\sqrt[3]{- 4 \sqrt{2} \left(1 - i\right)} = 2 \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

$\textcolor{w h i t e}{\sqrt[3]{- 4 \sqrt{2} \left(1 - i\right)}} = 2 \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\right)$

$\textcolor{w h i t e}{\sqrt[3]{- 4 \sqrt{2} \left(1 - i\right)}} = \sqrt{2} + \sqrt{2} i$

The other roots can be found by multiplying by $\omega$ or ${\omega}^{2} = \overline{\omega}$ where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive complex cube root of $1$.

$\left(\sqrt{2} + \sqrt{2} i\right) \left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

$= - \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} i - \frac{\sqrt{2}}{2} i - \frac{\sqrt{6}}{2}$

$= \frac{\sqrt{6} - \sqrt{2}}{2} + \frac{\sqrt{6} - \sqrt{2}}{2} i$

$\left(\sqrt{2} + \sqrt{2} i\right) \left(- \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)$

$= - \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2} i - \frac{\sqrt{2}}{2} i - \frac{\sqrt{6}}{2} i$

$= - \frac{\sqrt{2} + \sqrt{6}}{2} - \frac{\sqrt{2} + \sqrt{6}}{2} i$