What is the trigonometric form of (-2+9i) (2+9i)?

1 Answer
Jan 1, 2016

sqrt(85)(cos(tan^-1(-9/2))+isin(tan^-1(-9/2)))85(cos(tan1(92))+isin(tan1(92)))

Explanation:

(-2+9i)(2+9i)

rcos(theta)=-2rcos(θ)=2
rsin(thea) = 9rsin(thea)=9

Squaring both and adding

r^2cos^2(theta) = 4r2cos2(θ)=4
r^2sin^2(theta) = 81r2sin2(θ)=81

r^2cos^2(theta)+r^2sin^2(theta) = 4+81r2cos2(θ)+r2sin2(θ)=4+81
r^2(cos^2(theta)+sin^2(theta))=85r2(cos2(θ)+sin2(θ))=85
r^2=85r2=85
r=sqrt(85)r=85

rsin(theta)/rcos(theta) = 9/-2rsin(θ)rcos(θ)=92
tan(theta) = -9/2tan(θ)=92

theta = tan^-1(-9/2)θ=tan1(92)

The complex number in trigonometric form is

sqrt(85)(cos(tan^-1(-9/2))+isin(tan^-1(-9/2)))85(cos(tan1(92))+isin(tan1(92)))