# How do you divide ( - 6 i + 5) / (i+ 2 ) in trigonometric form?

$\frac{- 6 i + 5}{i + 2} = \sqrt{\frac{61}{5}} c i s \left(- {\tan}^{-} 1 \left(\frac{17}{-} 9\right)\right)$

#### Explanation:

$\frac{- 6 i + 5}{i + 2} = f \left(r , \theta\right)$
${z}_{1} = - 6 i + 5$
${r}_{1} = \sqrt{{\left(- 6\right)}^{2} + {5}^{2}} = \sqrt{36 + 25} = \sqrt{61}$
${\theta}_{1} = {\tan}^{-} 1 \left(\frac{5}{-} 6\right) = 2 \pi - {\tan}^{-} 1 \left(\frac{5}{6}\right)$
${z}_{2} = i + 2$
${r}_{2} = \sqrt{{1}^{2} + {2}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{2}{1}\right) = 2 \pi + {\tan}^{-} 1 2$
Thus,
$\frac{- 6 i + 5}{i + 2} = \frac{\sqrt{61} , \left(2 \pi - {\tan}^{-} 1 \left(\frac{5}{6}\right)\right)}{\sqrt{5} , \left(2 \pi + {\tan}^{-} 1 2\right)}$
By De-Moivre's Theorem,

$\frac{\sqrt{61} , \left(2 \pi - {\tan}^{-} 1 \left(\frac{5}{6}\right)\right)}{\sqrt{5} , \left(2 \pi + {\tan}^{-} 1 2\right)}$
$= \frac{\sqrt{61}}{\sqrt{5}} \times c i s \left(2 \pi - {\tan}^{-} 1 \left(\frac{5}{6}\right) - \left(2 \pi + {\tan}^{-} 1 2\right)\right)$

$= \sqrt{\frac{61}{5}} c i s \left(- {\tan}^{-} 1 \left(\frac{5}{6}\right) - {\tan}^{-} 1 2\right)$
$= \sqrt{\frac{61}{5}} c i s \left(- \left({\tan}^{-} 1 \left(\frac{5}{6}\right) + {\tan}^{-} 1 2\right)\right)$

${\tan}^{-} 1 \left(\frac{5}{6}\right) + {\tan}^{-} 1 2 = {\tan}^{-} 1 \left(\frac{\frac{5}{6} + 2}{1 - \frac{5}{6} \times 2}\right)$

$= {\tan}^{-} 1 \left(\frac{5 \times 1 + 2 \times 6}{1 \times 6 - 5 \times 3}\right) = {\tan}^{-} 1 \left(\frac{5 + 12}{6 - 15}\right)$

$= {\tan}^{-} 1 \left(\frac{17}{-} 9\right)$
Thus,
$\frac{- 6 i + 5}{i + 2} = \sqrt{\frac{61}{5}} c i s \left(- {\tan}^{-} 1 \left(\frac{17}{-} 9\right)\right)$