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

Oct 31, 2017

$\frac{9 + i}{- 2 + i} = 4.05 \left(\cos 3.72 + i \sin 3.72\right)$

#### Explanation:

$Z = \frac{9 + i}{- 2 + i} = \frac{\left(9 + i\right) \left(- 2 - i\right)}{\left(- 2 + i\right) \left(- 2 - i\right)}$

$= \frac{- 18 - 11 i - {i}^{2}}{{\left(- 2\right)}^{2} - {i}^{2}} = \frac{- 17 - 11 i}{4 + 1} = - \frac{17}{5} - \frac{11}{5} i$

= -3.4 -2.2i ; [i^2=-1]

Modulus $| Z | = r = \sqrt{{\left(- 3.4\right)}^{2} + {\left(- 2.2\right)}^{2}} = 4.05$ ;

$\tan \alpha = \frac{b}{a} = \frac{- 2.2}{-} 3.4 = 0.647 \therefore \alpha = {\tan}^{-} 1 \left(0.647\right) = 0.574$

$\theta$ is on $3 r d$ quadrant $\therefore \theta = \alpha + \pi = 0.574 + \pi = 3.72$;

$\theta$ expressed in radian.

Argument : $\theta = 3.72 \therefore$ In trigonometric form expressed as

$r \left(\cos \theta = i \sin \theta\right) = 4.05 \left(\cos 3.72 + i \sin 3.72\right) \therefore$

$\frac{9 + i}{- 2 + i} = 4.05 \left(\cos 3.72 + i \sin 3.72\right)$ [Ans]