# How do you convert 9 + 12i to polar form?

$15 \left(\cos \left(\setminus {\tan}^{- 1} \left(\frac{4}{3}\right)\right) + i \sin \left(\setminus {\tan}^{- 1} \left(\frac{4}{3}\right)\right)\right)$

#### Explanation:

$9 + 12 i = r \left(\setminus \cos \setminus \theta + i \setminus \sin \setminus \theta\right)$

Where,

$r = \setminus \sqrt{{9}^{2} + {12}^{2}} = 15$

$\setminus \theta = \setminus {\tan}^{- 1} \left(\frac{12}{9}\right) = \setminus {\tan}^{- 1} \left(\frac{4}{3}\right)$

$\setminus \therefore 9 + 12 i$

$= 15 \left(\cos \left(\setminus {\tan}^{- 1} \left(\frac{4}{3}\right)\right) + i \sin \left(\setminus {\tan}^{- 1} \left(\frac{4}{3}\right)\right)\right)$