How do you divide #( -i-9) / (i-2)# in trigonometric form?

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Answer 1 · 2018-07-27T02:08:28

#color(indigo)(=> 3.4 + 2.2 i, " I Quadrant"#

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

#z_1 = -9 - i, z_2 = 2 - i#

#r_1 = sqrt(-9^2 + -1^2)^2) = sqrt 82#

#theta_1 = tan ^-1 (-1/ -9) ~~ 186.3402^@ = , " III Quadrant"#

#r_2 = sqrt(2^2 + (-1)^2) = sqrt 5#

#theta_2 = tan ^-1 (1/ -2) ~~ 153.4349^@, " II Quadrant"#

#z_1 / z_2 = sqrt(82 / 5) (cos (186.3402 - 153.4349) + i sin (186.3402 - 153.4349))#

#color(indigo)(=> 3.4 + 2.2 i, " I Quadrant"#