How do you divide # (8+9i)/(1-3i) # in trigonometric form?

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Answer 1 · 2018-06-24T05:41:28

#color(violet)(=> 3.9115 ( -0.499 + i 0.8665)#

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

#z_1 = -8 + i 9, z_2 = 1 - i 3#

#r_1 = sqrt(8^2 + 9^2) = sqrt 153#

#theta_1 = tan ^ (-1) (9/8) = -tan *-1 (1.125) = 48.37 ^@#

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

#theta_2 = tan ^ (-3/ 1) = tan^-1 (-3) = -71.57^@ = 288.43^@, " IV Quadrant"#

#z_1 / z_2 = sqrt(153/10) (cos (48.37- 288.43) + i sin (48.37 - 288.43))#

#color(violet)(=> 3.9115 ( -0.499 + i 0.8665)#