How do you divide # (1-6i) / (5-2i) # in trigonometric form?

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Answer 1 · 2018-07-09T11:10:21

#color(green)(=> 0.5861 - 0.9656 i)#

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

#z_1 = 1 - 6 i, z_2 = 5 - 2 i#

#r_1 = sqrt(1^2 + -6^2) = sqrt 37#

#theta_1 = tan ^ (-1) (-6/1) ~~ 279.46 ^@, " IV Quadrant"#

#r_2 = sqrt(5^2 + (-2)^2) = sqrt 29#

#theta_2 = tan ^-1 (-2/ 5) ~~ 338.2^@, " IV Quadrant"#

#z_1 / z_2 = sqrt(37/29) (cos (279.46- 338.2) + i sin (279.46- 338.2))#

#color(green)(=> 0.5861 - 0.9656 i)#