Let's convert each to trigonometric form:
Numerator: 3i-8 = -8 + 3 i 3i−8=−8+3i
Magnitude: r_N = sqrt((-8)^2 + 3^2) = sqrt(73) rN=√(−8)2+32=√73
Angle: theta_N = arctan(3/-8) = -arctan(3/8) θN=arctan(3−8)=−arctan(38)
Denominator: -i + 7 = 7 - i −i+7=7−i
Magnitude: r_D = sqrt(7^2 + (-1)^2) = sqrt(50) = 5sqrt(2)rD=√72+(−1)2=√50=5√2
Angle: theta_D = arctan(-1 / 7) = -arctan(1/7) θD=arctan(−17)=−arctan(17)
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)) −8+3i7−i=rNeiθNrDeiθD=rNrDei(θN−θD)
By Euler's formula, we get
=sqrt(73/50) [cos(theta_N - theta_D) + i sin(theta_N - theta_D)] =√7350[cos(θN−θD)+isin(θN−θD)]
=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)] =√7350[cos(θN)cos(θD)+sin(θN)sin(θD)]+i√7350[sin(θN)cos(θD)−cos(θN)sin(θD)]
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.
