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

Apr 9, 2018

To convert to polar form we need to find modulus and argument of $- 3 - 6 i$

#### Explanation:

Modulus:
$| \setminus - 3 - 6 i | = \sqrt{{\left(- 3\right)}^{2} + {\left(- 6\right)}^{2}}$
$| \setminus - 3 - 6 i | = \sqrt{45}$

Argument:
$\theta = {\tan}^{-} 1 \left(\frac{6}{3}\right)$
$\theta = {\tan}^{-} 1 \left(2\right)$

Then $- 3 - 6 i$ can be rewritten as
$\sqrt{45} \left(\cos \left({\tan}^{-} 1 \left(2\right)\right) + i \sin \left({\tan}^{-} 1 \left(2\right)\right)\right)$

or

$\sqrt{45} c i s \left({\tan}^{-} 1 \left(2\right)\right)$