# How do you convert -9 - 9sqrt2 i  to polar form?

Aug 12, 2017

$- 9 - 9 \sqrt{2} i = 9 \sqrt{3} \left(\cos \theta + i s i n \theta\right)$, where $\theta = {\tan}^{- 1} \sqrt{2}$ and $\theta$ is in $Q 3$.

#### Explanation:

If a complex number $a + b i = r \cos \theta + i r \sin \theta$, $a = r \cos \theta$ and $b = r \sin \theta$ and squaring and adding ${a}^{2} + {b}^{2} = {r}^{2}$ or $r = \sqrt{{a}^{2} + {b}^{2}}$.

Hence, as we have $- 9 - 9 \sqrt{2} i$, $r = \sqrt{{\left(- 9\right)}^{2} + {\left(- 9 \sqrt{2}\right)}^{2}} = \sqrt{81 + 162} = \sqrt{243} = 9 \sqrt{3}$.

Hence $\cos \theta = - \frac{1}{\sqrt{3}}$ and $\sin \theta = - \sqrt{\frac{2}{3}}$ i.e. $\theta$ is in third quadrant and $\tan \theta = \sqrt{2}$

and $- 9 - 9 \sqrt{2} i = 9 \sqrt{3} \left(\cos \theta + i s i n \theta\right)$, where $\theta = {\tan}^{- 1} \sqrt{2}$ and $\theta$ is in $Q 3$.