Given:
Divide the complex numbers in trigonometric form:
color(blue)((12i-3)/(7i-2)12i−37i−2
Rewrite in color(green)(z=(a+bi)z=(a+bi) form:
color(blue)((-3+12i)/(-2+7i)−3+12i−2+7i
Multiply and divide by the complex conjugate of the denominator
rArr (-3+12i)/(-2+7i)*(-2-7i)/(-2-7i)⇒−3+12i−2+7i⋅−2−7i−2−7i
rArr {(-3+12i)(-2-7i)}/{(-2+7i)(-2-7i)⇒(−3+12i)(−2−7i)(−2+7i)(−2−7i)
rArr {(-3+12i)(-2-7i)}/{(-2)^2-(7i)^2)⇒(−3+12i)(−2−7i)(−2)2−(7i)2
Note that i = sqrt(-1) and i^2 = (-1)i=√−1andi2=(−1) in complex number arithmetic.
rArr (6+21i-24i-84i^2)/(4-(49i^2)⇒6+21i−24i−84i24−(49i2)
rArr (90-3i)/(4-49*(-1))⇒90−3i4−49⋅(−1)
rArr (90-3i)/(4+49)⇒90−3i4+49
rArr (90-3i)/53⇒90−3i53
rArr 90/53 -3/53i⇒9053−353i
Next, we will find the Polar form
Important:
For any complex number color(blue)(a+bia+bi,
Polar form is given by
color(blue)(r[cos(theta)+i sin(theta)],r[cos(θ)+isin(θ)], where
color(brown)(r = sqrt(a^2+b^2),r=√a2+b2, and
color(brown)(theta = tan^(-1)(b/a)θ=tan−1(ba)
We have, for a+bia+bi
a=(90/53) and b = (-3/53)a=(9053)andb=(−353)
r=sqrt((90/53)^2+(-3/53)^2)r=√(9053)2+(−353)2
r=sqrt(8100/2809+9/2809)r=√81002809+92809
r=sqrt(8109/2809)r=√81092809
r=sqrt((8109)/(53*53))r=√810953⋅53
r=sqrt((901*9)/(53*53))r=√901⋅953⋅53
r=sqrt(901/53*9/53)r=√90153⋅953
color(brown)(r=(3sqrt(901))/53r=3√90153
Also,
theta = tan^-1[(-3/53)/(90/53)]θ=tan−1[−3539053]
rArr theta = -tan^-1(1/30)⇒θ=−tan−1(130)
Add 2pi2π, since thetaθ 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.
