What is the trigonometric form of (-2+10i)(3+3i) (2+10i)(3+3i)?

1 Answer
Mar 28, 2018

12sqrt(13)(cos(tan^-1(-2/3))+isin(tan^-1(-2/3)))~~43.267(cos(-0.588)+isin(-0.588))1213(cos(tan1(23))+isin(tan1(23)))43.267(cos(0.588)+isin(0.588))

Explanation:

(-2+10i)(3+3i)=-6+30i-6i+30i^2=-36+24i(2+10i)(3+3i)=6+30i6i+30i2=36+24i

Trigonometric form: r(isin(theta)+cos(theta))r(isin(θ)+cos(θ))
Rectangular form: a+bia+bi
r=sqrt(a^2+b^2)=sqrt(36^2+24^2)~~43.267r=a2+b2=362+24243.267
theta=tan^-1(b/a)=tan^-1(24/-36)θ=tan1(ba)=tan1(2436)
=tan^-1(-2/3)~~-0.588 (radians)=tan1(23)0.588(radians)

12sqrt(13)(cos(tan^-1(-2/3))+isin(tan^-1(-2/3)))1213(cos(tan1(23))+isin(tan1(23)))