How do you divide ( 9i- 2) / (- 5 i+ 2 ) in trigonometric form?

Dec 20, 2016

$= \frac{1}{29} \left(41 + 8 i\right)$

Explanation:

Rewrite 9i-2 as -2+9i. Now square and the numbers -2 and 9, which would be 85. Its square root is$\sqrt{85}$. Multiply and divide the expression -2 +9i with $\sqrt{85}$ as follows:

$\sqrt{85} \left(- \frac{2}{\sqrt{85}} + i \frac{9}{\sqrt{85}}\right)$

If $\theta$ is some angle then let $\cos \theta = - \frac{2}{\sqrt{85}} \mathmr{and} \sin \theta = \frac{9}{\sqrt{85}}$

Thus $- 2 + 9 i = \sqrt{85} \left(\cos \theta + i \sin \theta\right)$

Like wise -5i +2 would be 2 -5i =$\sqrt{29} \left(\frac{2}{\sqrt{29}} - i \frac{5}{\sqrt{29}}\right)$

If $\phi$ is some angle, then let $\cos \phi = \frac{2}{\sqrt{29}} \mathmr{and} \sin \phi = - \frac{5}{\sqrt{29}}$

Thus $2 - 5 i = \sqrt{29} \left(\cos \phi - \sin \phi\right)$

(9i-2)/(-5i+2)= sqrt(85/29) (cos theta +isin theta)/(cosphi-isin phi)=sqrt(85/29) e^(itheta)/e^(-iphi) =sqrt(85/29) e^(i(theta +phi)

$= \sqrt{\frac{85}{29}} \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$

$= \sqrt{\frac{85}{29}} \left[\left(\cos \theta \cos \phi - \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi + \cos \theta \sin \phi\right)\right]$

$= \sqrt{\frac{85}{29}} \left[\frac{- 2}{\sqrt{85}} \cdot \frac{2}{\sqrt{29}} - \frac{9}{\sqrt{85}} \cdot \frac{- 5}{\sqrt{29}} + i \left(\frac{9}{\sqrt{85}} \cdot \frac{2}{\sqrt{29}} + \frac{- 2}{\sqrt{85}} \cdot \frac{- 5}{\sqrt{29}}\right)\right]$

$\sqrt{\frac{85}{29}} \left(\frac{41}{\sqrt{85} \sqrt{29}} + i \frac{28}{\sqrt{85} \sqrt{29}}\right)$
$= \frac{1}{29} \left(41 + 28 i\right)$