# How do you divide  (9+2i) / (5+i)  in trigonometric form?

Aug 14, 2018

$\sqrt{\frac{85}{26}} \left(\cos {1.22}^{o} + i \sin {1.22}^{o}\right)$

#### Explanation:

Use ${e}^{i \theta} = \cos \theta + i \sin \theta$

$\frac{9 + 2 i}{5 + i}$

=$\frac{a {e}^{i \alpha}}{b {e}^{i \beta}} = \left(\frac{a}{b}\right) \left({e}^{i \left(\alpha - \beta\right)}\right)$,

= a/b ( cos ( alpha - beta) + i sin ( alpha - beta )

where

$a = \sqrt{{9}^{2} + {2}^{2}} = \sqrt{85}$,

$b = \sqrt{{5}^{2} + 1} = \sqrt{26}$,

$\alpha = \arccos \left(\frac{9}{a}\right)$ and

$\beta = \arccos \left(\frac{5}{b}\right)$.

$\frac{9 + 2 i}{5 + i}$

= sqrt(85/26)( cos ( arccos (9/sqrt85) - arccos ( 5/sqrt26))

$+ i \sin \left(\arccos \left(\frac{9}{\sqrt{85}}\right) - \arccos \left(\frac{5}{\sqrt{26}}\right)\right)$

= sqrt(85/26) ( cos ( 12.53^o - 11.31^o)

 +i sin ( 12.53^o - 11.31^o) )#

$= \sqrt{\frac{85}{26}} \left(\cos {1.22}^{o} + i \sin {1.22}^{o}\right)$

This is very very close to the value

1/26 ( 47 + i )