# How do you convert ((2sqrt3) 2i)/(1-sqrt3i) to polar form?

Jun 30, 2016

$\frac{\left(2 \sqrt{3}\right) 2 i}{1 - \sqrt{3} i} = 2 \sqrt{3} \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right)$

#### Explanation:

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

Let us first rationalize it by multiplying numerator and denominator by complex conjugate of denominator. Then above becomes

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

= $\frac{4 \sqrt{3} i + 12 {i}^{2}}{{1}^{2} - {\left(\sqrt{3} i\right)}^{2}}$

= $\frac{4 \sqrt{3} i + 12 {i}^{2}}{{1}^{2} - \left(3 {i}^{2}\right)} = \frac{4 \sqrt{3} i + 12 {i}^{2}}{1 + 3}$

= $\frac{- 12 + 4 \sqrt{3} i}{4} = - 3 + \sqrt{3} i$

Now when a complex number $a + b i$ is written in polar form as $r \left(\cos \theta + i \sin \theta\right)$, $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = \arctan \left(\frac{b}{a}\right)$

Hence, here $r = \sqrt{{\left(- 3\right)}^{2} + {\left(\sqrt{3}\right)}^{2}} = \sqrt{9 + 3} = \sqrt{12} = 2 \sqrt{3}$

and $\theta = \arctan \left(\frac{\sqrt{3}}{- 3}\right) = \arctan \left(- \frac{1}{\sqrt{3}}\right) = - \frac{\pi}{6}$

Hence, $\frac{\left(2 \sqrt{3}\right) 2 i}{1 - \sqrt{3} i} = 2 \sqrt{3} \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right)$