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=tan−1(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+22etan−1(29)i=√85etan−1(29)i
5i+12=sqrt(12^2+5^2)e^(tan^-1(5/12)i)=13e^(tan^-1(5/12)i)5i+12=√122+52etan−1(512)i=13etan−1(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))√85etan−1(29)i13etan−1(512)i
=sqrt(85)/13 * e^(tan^-1(2/9)i)/e^(tan^-1(5/12)i)=√8513⋅etan−1(29)ietan−1(512)i
=sqrt(85)/13 * e^(tan^-1(2/9)i-tan^-1(5/12)i)=√8513⋅etan−1(29)i−tan−1(512)i (1)
=sqrt(85)/13 * e^(-tan^(-1)(21/118)i)=√8513⋅e−tan−1(21118)i (2)
(1) Exponent Properties e^a/e^b=e^(a-b)eaeb=ea−b
(2) Tangent Properties tan^-1(a)-tan^-1(b)=tan^-1((a-b)/(1+ab))tan−1(a)−tan−1(b)=tan−1(a−b1+ab) (derive by substituting a=tan(x), b=tan(y) into tangent addition formula)