# How do you divide (i-2) / (-5i-8) in trigonometric form?

Aug 13, 2018

$= \sqrt{\frac{5}{89}} \left(\cos {58.662}^{o} - i \sin {58.662}^{p}\right)$

#### Explanation:

Use $z = \left(x , y\right) = r \left(\cos \theta , \sin \theta\right) , r = \sqrt{{x}^{2} + {y}^{2}} \ge 0 ,$

$0 \le \theta = \arctan \left(\frac{y}{x}\right) , \in {Q}_{1}$ or $Q - 4$

$= \arccos \left(\frac{y}{r}\right) , \theta \in {Q}_{2}$

$= \pi + \arctan \left(\frac{y}{x}\right) , \theta \in {Q}_{3}$

The complex z = ( x + i y ) = r ( cos theta + i sin theta ) = r e^(i

theta ).

Here, z = u/v,

Q_2 u = - 2 + i = sqrt 5 e^(i cos^(-1)((-2)/sqrt 5) and

Q_3 v = - 8 - 5i = sqrt89e^(iarctan( 5/8). Now,

z = qsqrt(5/89)( e^(i cos^(-1)((-2)/sqrt 5))/(e^(i(pi +arctan( 5/8)))

$= \sqrt{\frac{5}{89}} {e}^{i} \left({\cos}^{- 1} \left(\frac{- 2}{\sqrt{5}}\right) - \left(\pi + \arctan \left(\frac{5}{8}\right)\right)\right)$

$= \sqrt{\frac{5}{89}} {e}^{i} \left({153.3435}^{o} - {180}^{o} - {32.0054}^{o}\right)$

$= \sqrt{\frac{5}{89}} {e}^{i} \left(- {58.662}^{o}\right)$

$= \sqrt{\frac{5}{89}} \left(\cos {58.662}^{o} - i \sin {58.662}^{p}\right)$

The answer is 1/89( 11- 18 i ).

My answer appears to be very very close.