How do you divide #(2i-7) / (-2i+6)# in trigonometric form?

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Answer 1 · 2018-07-08T02:43:43

#color(maroon)((-7 + 2i) / (6 - 2i) = -1.15 - 0.05i)#

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

#z_1 = -7 + 2i , z_2 = 6 - 2i #

#|r_1| = sqrt(-7^2 + 2^2) = sqrt 53#

#theta_1 = tan ^ (-1) (2/-7) = 164.05 ^@ " II Quadrant"#

#|r_2| = sqrt(6^2 + (2)^2) = sqrt 40#

#theta_2 = tan ^-1 (-2/ 6) = 341.57^@ , " IV Quadrant"#

#z_1 / z_2 = |sqrt(53/40)| * (cos (164.05- 341.57) + i sin (164.05- 341.57))#

#color(maroon)((-7 + 2i) / (6 - 2i) = -1.15 - 0.05i)#