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

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Answer 1 · 2018-07-04T16:33:22

#-0.623-0.694i#

#\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#

Answer 2 · 2018-07-08T01:41:08

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

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#