Let's convert each to trigonometric form:
Numerator: #3i-8 = -8 + 3 i #
Magnitude: #r_N = sqrt((-8)^2 + 3^2) = sqrt(73) #
Angle: #theta_N = arctan(3/-8) = -arctan(3/8) #
Denominator: #-i + 7 = 7 - i #
Magnitude: #r_D = sqrt(7^2 + (-1)^2) = sqrt(50) = 5sqrt(2)#
Angle: #theta_D = arctan(-1 / 7) = -arctan(1/7) #
Therefore, the first equation can be written as
#(-8+3i)/(7-i) = (r_Ne^(itheta_N))/(r_De^(itheta_D)) = r_N/r_D e^(i(theta_N - theta_D)) #
By Euler's formula, we get
#=sqrt(73/50) [cos(theta_N - theta_D) + i sin(theta_N - theta_D)] #
#=sqrt(73/50) [cos(theta_N)cos(theta_D) + sin(theta_N)sin(theta_D)] + i sqrt(73/50)[sin(theta_N)cos(theta_D) - cos(theta_N)sin(theta_D)] #
We know the cosines and sines of these angles based on the original numbers:
#cos(theta_N) = (-8)/sqrt(73)\ \ \ \ \ \ sin(theta_N) = 3/sqrt(73)#
#cos(theta_D) = 7/sqrt(50)\ \ \ \ \ \ sin(theta_D) = (-1)/sqrt(50)#
Hence,
#= sqrt(73/50)[(-8)/sqrt(73) * 7/(sqrt(50)) + 3/sqrt(73) * (-1)/sqrt(50)] + i sqrt(73/50) [3/sqrt(73) * 7/sqrt(50) + (-8)/sqrt(73) * (-1)/sqrt(50) ] #
#= 1/50[ -56 - 3 ]+ i/50[21 + 8] = (-59)/50 + 29/50 i#
We can check this with a Cartesian calculation:
#(-8+3i)/(7-i) * (7+i)/(7+i) = (-56 + 21 i - 3 + 8i)/50 = (-59)/50 + 29/50 i #
as expected.
