# How do you divide ( 9i- 7) / (- 5 i -6 ) in trigonometric form?

Nov 8, 2017

$\frac{\sqrt{130}}{\sqrt{61}} \cos \left(- 91.931\right) + i \sin \left(- 91.931\right)$

#### Explanation:

$z = a + b i$ in trigonometric form is $z = r \left(\cos \setminus \theta + i \sin \setminus \theta\right)$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\setminus \theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

So, we have $\frac{- 7 + 9 i}{- 6 - 5 i} = \frac{\sqrt{{\left(- 7\right)}^{2} + {9}^{2}} \left(\cos \left({\tan}^{- 1} \left(\frac{9}{-} 7\right)\right) + i \sin \left({\tan}^{- 1} \left(\frac{9}{-} 7\right)\right)\right)}{\sqrt{{\left(- 6\right)}^{2} + {\left(- 5\right)}^{2}} \left(\cos \left({\tan}^{- 1} \left(\frac{- 5}{-} 6\right)\right) + i \sin \left({\tan}^{- 1} \left(\frac{- 5}{-} 6\right)\right)\right)}$
$= \frac{\sqrt{130} \left(\cos \left(- 52.125\right) + i \sin \left(- 52.125\right)\right)}{\sqrt{61} \left(\cos \left(39.806\right) + i \sin \left(39.806\right)\right)}$
$= \frac{\sqrt{130}}{\sqrt{61}} \cos \left(- 52.125 - 39.806\right) + i \sin \left(- 52.125 - 39.806\right)$
$= \frac{\sqrt{130}}{\sqrt{61}} \cos \left(- 91.931\right) + i \sin \left(- 91.931\right)$