How do you divide # (-4+2i)/(3-i) # in trigonometric form?

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Answer 1 · 2018-07-02T11:11:37

#color(maroon)((-4 + 2i) / (3 - i) ~~ -1.4 - 0.2 i#

To divide #(-4 + 2 i) / (3 - i)# using trigonometric form.

#z_1 = (-4 + 2 i), z_2 = (3 - i)#

#r_1 = sqrt(-4^2 + 2^2) = sqrt 20

#r_2 = sqrt(3^2 + -1^2) = sqrt 10#

#theta_1 = arctan (2/-4) = 153.43^@, " II quadrant"#

#Theta_2 = arctan(-1/3) = 341.57^@, " IV 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(10/20) * (cos (153.43 - 341.57 ) + i sin (153.43 - 341.57 )#

#z_1 / z_2 = sqrt(20/10) * (cos (-188.14) + i sin (-188.14))#

#color(maroon)((-4 + 2i) / (3 - i) ~~ -1.4 - 0.2 i#