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)