How do you divide ( 2i+9) / ( 5 i +12 )2i+95i+12 in trigonometric form?

1 Answer
Jun 5, 2018

sqrt(85)/13 * e^(-tan^(-1)(21/118)i)8513etan1(21118)i

with r=sqrt(85)/13,theta=-tan^-1(21/118)r=8513,θ=tan1(21118)

Explanation:

First, express (2i+9),(5i+12)(2i+9),(5i+12) in polar form.

Converting to Polar Form

For a complex number a+bia+bi, the polar form is r(costheta+isintheta)=re^(itheta)r(cosθ+isinθ)=reiθ (Euler's Formula), hence

rcostheta=arcosθ=a (Equation 1)

rsintheta=brsinθ=b (Equation 2)

Squaring both equations and adding them together, we have

r^2(cos^2theta+sin^2theta)=a^2+b^2r2(cos2θ+sin2θ)=a2+b2

Since cos^2theta+sin^2theta=1cos2θ+sin2θ=1

r^2=a^2+b^2r2=a2+b2

color(red)(r=sqrt(a^2+b^2))r=a2+b2

To solve for thetaθ, we can divide Equation 2 by Equation 1, yielding

(rsintheta)/(rcostheta)=tantheta=b/arsinθrcosθ=tanθ=ba

Hence color(red)(theta=tan^-1(b/a))θ=tan1(ba)

Substituting (2i+9),(5i+12)(2i+9),(5i+12) for a+bia+bi, we have their polar forms

2i+9=sqrt(9^2+2^2)e^(tan^-1(2/9)i)=sqrt(85)e^(tan^-1(2/9)i)2i+9=92+22etan1(29)i=85etan1(29)i

5i+12=sqrt(12^2+5^2)e^(tan^-1(5/12)i)=13e^(tan^-1(5/12)i)5i+12=122+52etan1(512)i=13etan1(512)i

Division in Polar Form

Now (2i+9)/(5i+12)2i+95i+12 becomes

(sqrt(85)e^(tan^-1(2/9)i))/(13e^(tan^-1(5/12)i))85etan1(29)i13etan1(512)i

=sqrt(85)/13 * e^(tan^-1(2/9)i)/e^(tan^-1(5/12)i)=8513etan1(29)ietan1(512)i

=sqrt(85)/13 * e^(tan^-1(2/9)i-tan^-1(5/12)i)=8513etan1(29)itan1(512)i (1)

=sqrt(85)/13 * e^(-tan^(-1)(21/118)i)=8513etan1(21118)i (2)

(1) Exponent Properties e^a/e^b=e^(a-b)eaeb=eab

(2) Tangent Properties tan^-1(a)-tan^-1(b)=tan^-1((a-b)/(1+ab))tan1(a)tan1(b)=tan1(ab1+ab) (derive by substituting a=tan(x), b=tan(y) into tangent addition formula)