# How do you divide ( 2i -4) / ( 5 i -6 ) in trigonometric form?

Feb 18, 2018

(2i-4)/(5i-6)=sqrt(5/61)"cis(tan^-1(-16/29))

#### Explanation:

$\text{Let" z_1=2i-4, "where} ,$
${r}_{1} = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{- 4}{2}\right)$

$\text{Let" z_2=5i-6, "where} ,$
${r}_{1} = \sqrt{{5}^{2} + {\left(- 6\right)}^{2}} = \sqrt{61}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{- 6}{5}\right)$

$\text{Now,}$

${z}_{1} = \sqrt{5} \left(\text{cis} \left({\tan}^{-} 1 \left(\frac{- 4}{2}\right)\right)\right)$

${z}_{2} = \sqrt{61} \left(\text{cis} \left({\tan}^{-} 1 \left(\frac{- 6}{5}\right)\right)\right)$

$\frac{2 i - 4}{5 i - 6} = {z}_{1} / {z}_{2} = \left(\sqrt{5} \left(\text{cis"(tan^-1((-4)/2))))/(sqrt61("cis} \left({\tan}^{-} 1 \left(\frac{- 6}{5}\right)\right)\right)\right)$
$= \frac{\sqrt{5}}{\sqrt{61}} \left(\text{cis"(tan^-1((-4)/2)))/("cis} \left({\tan}^{-} 1 \left(\frac{- 6}{5}\right)\right)\right)$

$\frac{\sqrt{5}}{\sqrt{61}} = \sqrt{\frac{5}{61}}$
$\text{By De-Moivre's Theorem,}$

$\left(\text{cis"(tan^-1((-4)/2)))/("cis} \left({\tan}^{-} 1 \left(\frac{- 6}{5}\right)\right)\right) = c i s \left({\tan}^{-} 1 \left(\frac{- 4}{2}\right) - {\tan}^{-} 1 \left(\frac{- 6}{5}\right)\right)$

${\tan}^{-} 1 \left(\frac{- 4}{2}\right) - {\tan}^{-} 1 \left(\frac{- 6}{5}\right) = {\tan}^{-} 1 \left(\frac{\frac{- 4}{2} - \frac{- 6}{5}}{1 + \left(\frac{- 4}{2}\right) \left(\frac{- 6}{5}\right)}\right)$

$- \frac{4}{2} - \frac{6}{5} = - 2 - \frac{6}{5} = \frac{- 10 - 6}{5} = - \frac{16}{5}$

$1 + \left(\frac{- 4}{2}\right) \left(\frac{- 6}{5}\right) = 1 + \frac{24}{5} = \frac{5 + 24}{5} = \frac{29}{5}$

Hence,
${\tan}^{-} 1 \left(\frac{\frac{- 4}{2} - \frac{- 6}{5}}{1 + \left(\frac{- 4}{2}\right) \left(\frac{- 6}{5}\right)}\right) = {\tan}^{-} 1 \left(\frac{- \frac{16}{5}}{\frac{29}{5}}\right) = {\tan}^{-} 1 \left(- \frac{16}{29}\right)$

Thus,

(2i-4)/(5i-6)=sqrt(5/61)"cis(tan^-1(-16/29))