How do you divide ( i+2) / (7i +3) in trigonometric form?

Jan 5, 2018

$\frac{\sqrt{290}}{58} \left(\cos \left(40.24\right) - i \sin \left(40.24\right)\right)$

Explanation:

For a complex number $z = a + b i$, we can rewrite it in the form $z = r \left(\cos \theta + i \sin \theta\right)$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

${z}_{1} = 2 + i$
${r}_{1} = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$
${\theta}_{1} = {\tan}^{- 1} \left(\frac{1}{2}\right)$
z_1=sqrt(5)(cos(tan^(-1)(1/2))+isin(tan^(-1)(1/2))

${z}_{2} = 3 + 7 i$
${r}_{2} = \sqrt{{3}^{2} + {7}^{2}} = \sqrt{58}$
${\theta}_{2} = {\tan}^{- 1} \left(\frac{7}{3}\right)$
z_2=sqrt(5)(cos(tan^(-1)(7/3))+isin(tan^(-1)(7/3))

${z}_{1} / {z}_{2} = {r}_{1} / {r}_{2} \left(\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right)$

${z}_{1} / {z}_{2} = \frac{\sqrt{5}}{\sqrt{58}} \left(\cos \left({\tan}^{- 1} \left(\frac{1}{2}\right) - {\tan}^{- 1} \left(\frac{7}{3}\right)\right) + i \sin \left({\tan}^{- 1} \left(\frac{1}{2}\right) - {\tan}^{- 1} \left(\frac{7}{3}\right)\right)\right)$
$\approx \frac{\sqrt{290}}{58} \left(\cos \left(- 40.24\right) + i \sin \left(- 40.24\right)\right)$

Since $\cos \left(x\right) = \cos \left(- x\right)$ and $\sin \left(- x\right) = - \sin \left(x\right)$

$= \frac{\sqrt{290}}{58} \left(\cos \left(40.24\right) - i \sin \left(40.24\right)\right)$