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

1 Answer
Aug 13, 2018

= sqrt( 5/89)( cos 58.662^o - i sin 58.662^p )

Explanation:

Use z = ( x, y ) = r ( cos theta, sin theta ), r = sqrt ( x^2 + y^2 )>= 0,

0 <= theta = arctan ( y/x),in Q_1 or Q-4

= arccos ( y/r), theta in Q_2

= pi + arctan(y/x), 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( 5/89) e^i ( cos^(-1)((-2)/sqrt 5) - (pi +arctan( 5/8)))

= sqrt( 5/89) e^i(153.3435^o - 180^o - 32.0054^o)

= sqrt( 5/89) e^i(-58.662^o)

= sqrt( 5/89)( cos 58.662^o - i sin 58.662^p )

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

My answer appears to be very very close.