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

1 Answer
Jun 5, 2018

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

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

Explanation:

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

Converting to Polar Form

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

rcostheta=a (Equation 1)

rsintheta=b (Equation 2)

Squaring both equations and adding them together, we have

r^2(cos^2theta+sin^2theta)=a^2+b^2

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

r^2=a^2+b^2

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

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

(rsintheta)/(rcostheta)=tantheta=b/a

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

Substituting (2i+9),(5i+12) for a+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)

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

Division in Polar Form

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

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

=sqrt(85)/13 * e^(tan^-1(2/9)i)/e^(tan^-1(5/12)i)

=sqrt(85)/13 * e^(tan^-1(2/9)i-tan^-1(5/12)i) (1)

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

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

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