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

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Answer 1 · 2018-07-27T02:08:28

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

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

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

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

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

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

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

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

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

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