Given:
Divide the complex numbers in trigonometric form:
#color(blue)((12i-3)/(7i-2)#
Rewrite in #color(green)(z=(a+bi)# form:
#color(blue)((-3+12i)/(-2+7i)#
Multiply and divide by the complex conjugate of the denominator
#rArr (-3+12i)/(-2+7i)*(-2-7i)/(-2-7i)#
#rArr {(-3+12i)(-2-7i)}/{(-2+7i)(-2-7i)#
#rArr {(-3+12i)(-2-7i)}/{(-2)^2-(7i)^2)#
Note that #i = sqrt(-1) and i^2 = (-1)# in complex number arithmetic.
#rArr (6+21i-24i-84i^2)/(4-(49i^2)#
#rArr (90-3i)/(4-49*(-1))#
#rArr (90-3i)/(4+49)#
#rArr (90-3i)/53#
#rArr 90/53 -3/53i#
Next, we will find the Polar form
Important:
For any complex number #color(blue)(a+bi#,
Polar form is given by
#color(blue)(r[cos(theta)+i sin(theta)],# where
#color(brown)(r = sqrt(a^2+b^2),# and
#color(brown)(theta = tan^(-1)(b/a)#
We have, for #a+bi#
#a=(90/53) and b = (-3/53)#
#r=sqrt((90/53)^2+(-3/53)^2)#
#r=sqrt(8100/2809+9/2809)#
#r=sqrt(8109/2809)#
#r=sqrt((8109)/(53*53))#
#r=sqrt((901*9)/(53*53))#
#r=sqrt(901/53*9/53)#
#color(brown)(r=(3sqrt(901))/53#
Also,
#theta = tan^-1[(-3/53)/(90/53)]#
#rArr theta = -tan^-1(1/30)#
Add #2pi#, since #theta# is negative.
#color(brown)[:. theta = -tan^-1(1/30)+2pi#
Hence, the final answer is given by
#color(blue)(90/53-(3i)/53 = (3/53)sqrt(901){cos(-a tan(1/30)+2pi}+i sin{-a tan(1/30)+2pi}#
Hope you find this solution useful.
