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

1 Answer
Martin C.
Mar 28, 2018

12sqrt(13)(cos(tan^-1(-2/3))+isin(tan^-1(-2/3)))~~43.267(cos(-0.588)+isin(-0.588))12√13(cos(tan−1(−23))+isin(tan−1(−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+30i−6i+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+242≈43.267
theta=tan^-1(b/a)=tan^-1(24/-36)θ=tan−1(ba)=tan−1(24−36)
=tan^-1(-2/3)~~-0.588 (radians)=tan−1(−23)≈−0.588(radians)

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

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