# How do you divide  (1-3i) / (-2+i)  in trigonometric form?

Feb 24, 2018

$\frac{1 - 3 i}{- 2 + i} = \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

#### Explanation:

For the complex numbers $z$ and $w$,

$| \frac{z}{w} | = | z \frac{|}{|} w |$

and

$a r g \left(\frac{z}{w}\right) = a r g \left(z\right) - a r g \left(w\right)$

So let $z = 1 - 3 i$ and let $w = - 2 + i$

$| z | = \sqrt{{1}^{2} + {3}^{2}}$
$= \sqrt{10}$

$\tan \theta = \frac{3}{1}$
$\theta = {\tan}^{-} 1 \left(3\right)$
$\therefore a r g \left(z\right) = - {\tan}^{-} 1 \left(3\right)$

$| w | = \sqrt{{2}^{2} + {1}^{2}}$
$= \sqrt{5}$

$\tan \theta = \frac{1}{2}$
$\theta = {\tan}^{-} 1 \left(\frac{1}{2}\right)$
$\therefore a r g \left(w\right) = \pi - {\tan}^{-} 1 \left(\frac{1}{2}\right)$

$\therefore | \frac{z}{w} | = \frac{\sqrt{10}}{\sqrt{5}}$
$= \sqrt{\frac{10}{5}}$
$= \sqrt{2}$

$a r g \left(\frac{z}{w}\right) = - {\tan}^{-} 1 \left(3\right) - \left(\pi - {\tan}^{-} 1 \left(\frac{1}{2}\right)\right)$
$= {\tan}^{-} 1 \left(\frac{1}{2}\right) - {\tan}^{-} 1 \left(3\right) - \pi$
$= - \frac{5}{4} \pi$

since $- \pi < a r g \left(\frac{z}{w}\right) \le \pi$

$a r g \left(\frac{z}{w}\right) = \frac{3 \pi}{4}$

$\therefore \frac{z}{w} = \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$