How do you cube 1-sqrt3i?

Oct 15, 2015

-8

Explanation:

You will have to multiply it one at a time.

${\left(1 - i \sqrt{3}\right)}^{3}$

$= \left(1 - i \sqrt{3}\right) \left(1 - i \sqrt{3}\right) \left(1 - i \sqrt{3}\right)$

$= \left(1 - i \sqrt{3}\right) \left(1 - i \sqrt{3} - i \sqrt{3} - 3\right)$

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

$= 1 - 2 i \sqrt{3} - 3 - i \sqrt{3} - 6 + 3 i \sqrt{3}$

$= \textcolor{red}{- 8}$

Oct 15, 2015

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

So ${\left(1 - \sqrt{3} i\right)}^{3} = {2}^{3} \left(\cos \left(- \pi\right) + i \sin \left(- \pi\right)\right) = {2}^{3} \cdot - 1 = - 8$

Explanation:

De Moivre's formula tells us that:

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

Now

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

So

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

$= {2}^{3} {\left(\cos \left(- \frac{\pi}{3}\right) + i \sin \left(- \frac{\pi}{3}\right)\right)}^{3} = 8 \left(\cos \left(- \pi\right) + i \sin \left(- \pi\right)\right)$

$= 8 \cdot - 1 = - 8$