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.
