How do you divide ( -i+2) / (i +4 ) in trigonometric form?

Oct 21, 2016

$= \frac{1}{17} \left(7 - 6 i\right)$

Explanation:

In trig form we have

$\frac{{R}_{1} {e}^{i {\theta}_{1}}}{{R}_{2} {e}^{i {\theta}_{2}}}$

$= \frac{{R}_{1}}{{R}_{2}} {e}^{i \left({\theta}_{1} - {\theta}_{2}\right)}$

${R}_{1} = \sqrt{{\left(2\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$

${R}_{2} = \sqrt{{4}^{2} + {1}^{2}} = \sqrt{17}$

$\tan {\theta}_{1} = - \frac{1}{2}$

$\tan {\theta}_{2} = \frac{1}{4}$

From
$\tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

$\tan \left({\theta}_{1} - {\theta}_{2}\right) = \frac{\left(- \frac{1}{2}\right) - \frac{1}{4}}{1 + \left(- \frac{1}{2}\right) \frac{1}{4}} = - \frac{6}{7}$

Which means that $\cos \left({\theta}_{1} - {\theta}_{2}\right) = \frac{7}{\sqrt{85}}$ and $\sin \left({\theta}_{1} - {\theta}_{2}\right) = - \frac{6}{\sqrt{85}}$

So

$\frac{{R}_{1} {e}^{i {\theta}_{1}}}{{R}_{2} {e}^{i {\theta}_{2}}}$

$= \frac{\sqrt{5}}{\sqrt{17}} {e}^{i \left(\arctan - \frac{6}{7}\right)}$

$= \frac{\sqrt{5}}{\sqrt{17}} \left(\frac{7}{\sqrt{85}} - i \frac{6}{\sqrt{85}}\right)$

$= \frac{1}{17} \left(7 - 6 i\right)$

It's a lot simpler by finding the complex conjugate of the denominator ${d}^{p} r i m e$ and multiply the whole thing by the by $\frac{{d}^{\setminus} p r i m e}{{d}^{p} r i m e}$. As follows

$\frac{2 - i}{4 + i} \cdot \frac{4 - i}{4 - i}$

$= \frac{8 - 2 i - 4 i - 1}{16 - 4 i + 4 i + 1}$

$= \frac{7 - 6 i}{17}$

$= \frac{1}{17} \left(7 - 6 i\right)$