# How do you convert sqrt3 - 3i to polar form?

Mar 21, 2018

The answer is $= \sqrt{12} \left(\cos \left(- \frac{1}{3} \pi\right) + i \sin \left(- \frac{1}{3} \pi\right)\right)$

#### Explanation:

Any complex number

$z = a + i b$

can be converted to the polar form

$z = | z | \left(\cos \theta + i \sin \theta\right)$

Where,

$\cos \theta = \frac{a}{| z |}$

and

$\sin \theta = \frac{b}{| z |}$

Here,

$z = \sqrt{3} - 3 i$

$| z | = \sqrt{{\left(\sqrt{3}\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{12}$

$\cos \theta = \frac{\sqrt{3}}{\sqrt{12}} = \frac{1}{2}$

$\sin \theta = - \frac{3}{\sqrt{12}} = - \frac{\sqrt{3}}{2}$

Therefore,

$\theta = - \frac{\pi}{3}$, $\left[\mod 2 \pi\right]$

$z = \sqrt{12} \left(\cos \left(- \frac{1}{3} \pi\right) + i \sin \left(- \frac{1}{3} \pi\right)\right)$