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

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Answer 1 · 2018-08-13T04:22:26

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

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.