Use z = ( x, y ) = r ( cos theta, sin theta ), r = sqrt ( x^2 + y^2 )>= 0,z=(x,y)=r(cosθ,sinθ),r=√x2+y2≥0,
0 <= theta = arctan ( y/x),in Q_10≤θ=arctan(yx),∈Q1 or Q-4Q−4
= arccos ( y/r), theta in Q_2=arccos(yr),θ∈Q2
= pi + arctan(y/x), theta in Q_3=π+arctan(yx),θ∈Q3
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)Q2u=−2+i=√5eicos−1(−2√5) and
Q_3 v = - 8 - 5i = sqrt89e^(iarctan( 5/8)Q3v=−8−5i=√89eiarctan(58). 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)))=√589ei(cos−1(−2√5)−(π+arctan(58)))
= sqrt( 5/89) e^i(153.3435^o - 180^o - 32.0054^o)=√589ei(153.3435o−180o−32.0054o)
= sqrt( 5/89) e^i(-58.662^o)=√589ei(−58.662o)
= sqrt( 5/89)( cos 58.662^o - i sin 58.662^p )=√589(cos58.662o−isin58.662p)
The answer is 1/89( 11- 18 i ).
My answer appears to be very very close.
