# How do you convert -6 - 6i sqrt 3 to polar form?

Jul 14, 2016

12(sin ((-5pi)/6)) - i cos ((-5pi)/6))

#### Explanation:

In polar form,$\left(r \sin \theta + r i \cos \theta\right)$ ... $I$
Comparing it with $- 6 - 6 i \sqrt{3}$
we get $\rightarrow - 6 = r \sin \theta$ ... 1
$\rightarrow - 6 \sqrt{3} = r \cos \theta$ ... 2

Squaring and adding 1 and 2,
${r}^{2} {\sin}^{2} \theta + {r}^{2} {\cos}^{\theta} = {\left(- 6\right)}^{2} + {\left(- 6 \sqrt{3}\right)}^{2}$
${r}^{2} \left({\sin}^{2} \theta + {\cos}^{2} \theta\right) = 36 + 108$
${r}^{2} = 144$
$r = 12$

Dividing equation 1 by 2
$\tan \theta = \frac{- 6}{- 6 \sqrt{3}}$
$\tan \theta = \frac{1}{\sqrt{3}}$
$\theta = \frac{\pi}{6}$

Real part of complex no = $- 6 \left(- x\right)$
Imaginary part of complex no = $- 6 \sqrt{3} \left(- y\right)$
$\therefore$the point is in 3rd quadrant.

At 3rd quadrant $\alpha = \theta - \pi$
$\alpha = \frac{- 5 \pi}{6}$

Substituting the value in equation $I$ we get,
12(sin ((-5pi)/6)) - i cos ((-5pi)/6))