How do you divide ( 3i-8) / (-i +7 ) in trigonometric form?

1 Answer
Mar 31, 2018

(-59)/50 + 29/50 i

Explanation:

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.