How do you cube 1+sqrt3i?

Oct 19, 2015

Use Binomial expansion or De Moivre's formula...

Explanation:

Method 1

Binomial expansion...

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

$= 1 + 3 \sqrt{3} i + 3 {\sqrt{3}}^{2} {i}^{2} + {\sqrt{3}}^{3} {i}^{3}$

$= 1 + 3 \sqrt{3} i + 9 {i}^{2} + 3 \sqrt{3} {i}^{3}$

$= 1 + 3 \sqrt{3} i - 9 - 3 \sqrt{3} i$

$= \left(1 - 9\right) + \left(3 \sqrt{3} - 3 \sqrt{3}\right) i$

$= - 8$

Method 2

Using De Moivre's formula...

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

So

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

$= 8 \left(\cos \pi + i \sin \pi\right) = 8 \cdot - 1 = - 8$