# How do you divide  (7-9i)/(-2-9i)  in trigonometric form?

$\frac{\sqrt{442}}{17} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 81}{-} 67\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(\frac{- 81}{-} 67\right)\right)\right]$ OR

$\frac{\sqrt{442}}{17} \left[\cos \left({50.403791360249}^{\circ}\right) + i \cdot \sin \left({50.403791360249}^{\circ}\right)\right]$

#### Explanation:

Convert to Trigonometric forms first

$7 - 9 i = \sqrt{130} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 9}{7}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 9}{7}\right)\right)\right]$

$- 2 - 9 i = \sqrt{85} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 9}{-} 2\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 9}{-} 2\right)\right)\right]$

Divide equals by equals

$\frac{7 - 9 i}{- 2 - 9 i} =$

$\left(\frac{\sqrt{130}}{\sqrt{85}}\right) \left[\cos \left({\tan}^{-} 1 \left(\frac{- 9}{7}\right) - {\tan}^{-} 1 \left(\frac{- 9}{-} 2\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 9}{7}\right) - {\tan}^{-} 1 \left(\frac{- 9}{-} 2\right)\right)\right]$

Take note of the formula:

$\tan \left(A - B\right) = \frac{T a n A - T a n B}{1 + T a n A \cdot T a n B}$

also

$A - B = T a {n}^{-} 1 \left(\frac{T a n A - T a n B}{1 + T a n A \cdot T a n B}\right)$

$\frac{\sqrt{442}}{17} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 81}{-} 67\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(\frac{- 81}{-} 67\right)\right)\right]$

have a nice day!