# How do you divide  (8-2i) / (5-3i)  in trigonometric form?

sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23)) OR

$\sqrt{2} \left(\cos \left({16.9275}^{\circ}\right) + i \sin \left({16.9275}^{\circ}\right)\right)$

#### Explanation:

Start from the given complex number

$\frac{8 - 2 i}{5 - 3 i}$

Convert the numerator

${r}_{1} = \sqrt{{8}^{2} + {\left(- 2\right)}^{2}} = \sqrt{68}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{- 2}{8}\right) = {\tan}^{-} 1 \left(\frac{- 1}{4}\right)$

then

$8 - 2 i = \sqrt{68} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right)\right]$

Convert the denominator

${r}_{2} = \sqrt{{5}^{2} + {\left(- 3\right)}^{2}} = \sqrt{34}$

${\theta}_{2} = {\tan}^{-} 1 \left(\frac{- 3}{5}\right) = {\tan}^{-} 1 \left(\frac{- 3}{5}\right)$

then

$5 - 3 i = \sqrt{34} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right)\right]$

Let us divide now, from the given with the equivalent

$\frac{8 - 2 i}{5 - 3 i} = \frac{\sqrt{68} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right)\right]}{\sqrt{34} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right)\right]}$

Divide using the following formula

$\left({r}_{1} / {r}_{2}\right) \cdot \left[\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left(\theta 1 - {\theta}_{2}\right)\right]$

(8-2i)/(5-3i)= sqrt(68/34)[cos(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))+isin(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))]

Take note: that
$\tan \left({\theta}_{1} - {\theta}_{2}\right) = \frac{\tan {\theta}_{1} - \tan {\theta}_{2}}{1 + \tan {\theta}_{1} \cdot \tan {\theta}_{2}} = \frac{- \frac{1}{4} - \left(- \frac{3}{5}\right)}{1 + \left(- \frac{1}{4}\right) \cdot \left(- \frac{3}{5}\right)}$

$\tan \left({\theta}_{1} - {\theta}_{2}\right) = \frac{7}{23}$

${\theta}_{1} - {\theta}_{2} = {\tan}^{-} 1 \left(\frac{7}{23}\right)$

and

(8-2i)/(5-3i)=sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23))

Have a nice day!!!.