How do you divide ( -7i-5) / ( 9 i -2 ) in trigonometric form?

2 Answers

-0.623-0.694i

Explanation:

\frac{-7i-5}{9i-2}

=\frac{-5-7i}{-2+9i}

=\frac{\sqrt{(-5)^2+(-7)^2}\ e^{i(-\pi+\tan^{-1}(7/5))}}{\sqrt{(-2)^2+(9)^2}\ e^{i(\pi-\tan^{-1}(9/2))}}

=\frac{\sqrt{74}\ e^{i(-\pi+\tan^{-1}(7/5))}}{\sqrt{85}\ e^{i(\pi-\tan^{-1}(9/2))}}

=\sqrt{74/85}e^{i(-\pi+\tan^{-1}(7/5))-i(\pi-\tan^{-1}(9/2))}

=\sqrt{74/85}e^{i(-2\pi+\tan^{-1}(7/5)+\tan^{-1}(9/2))}

=\sqrt{74/85}e^{-3.9805i)

=\sqrt{74/85}(\cos(-3.9805)+i\sin(-3.9805))

=\sqrt{74/85}(\cos(3.9805)-i\sin(3.9805))

=-0.623-0.694i

Jul 8, 2018

color(chocolate)((-5 - 7 i) / (-2 +9 i) ~~ 0.6235 + i 0.6942

Explanation:

To divide (-5 - 7 i) / (-2 +9 i) using trigonometric form.

z_1 = (-5 - 7 i), z_2 = (-2 +9 i)

#r_1 = sqrt(-5^2 - 7^2) = sqrt 74

r_2 = sqrt(-2^2 + 9^2) = sqrt 85

theta_1 = arctan (-5/-7) = 215.54^@, " III quadrant"

Theta_2 = arctan(-2/9) = 167.47^@, " II quadrant"

z_1 / z_2 = (r_1 / r_2) * (cos (theta_1 - theta_2) + i sin (theta_1 - theta_2))

z_1 / z_2 = sqrt(74/85) * (cos (215.54 - 167.47 ) + i sin (215.54 - 167.47 ))

z_1 / z_2 = sqrt(74/85) * (cos (48.07) + i sin (48.07))

color(chocolate)((-5 - 7 i) / (-2 +9 i) ~~ 0.6235 + i 0.6942