How do you divide #( - 6 i + 5) / (-2i+6)# in trigonometric form?

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Answer 1 · 2018-06-15T04:43:05

#color(violet)(=> 1.235 (0.8503 - i 0.526)#

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

#z_1 = 5 - i6, z-2 = 6 - i2#

#r_1 = sqrt(5^2 + 6^2) = sqrt61#

#theta_1 = arctan (-6/5) = 309.81^@, " IV Quadrant"#

#r_2 = sqrt(6^2 + 2^2) = sqrt40#

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

#z_1 / z_2 = (sqrt(61/40) (cos (309.81 - 341.57) + i sin(129.81 - 161.57)#

#color(violet)(=> 1.235 (0.8503 - i 0.5264)#