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)
