How do you divide $(2-9i)/(6+2i)$ in trigonometric form?

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Answer 1 · 2018-06-24T03:41:28

$color(crimson)(=> 1.33 ( -0.1028 - i 0.9947)$

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

$z_1 = 2 - i 9, z_2 = 6 + i 2$

$r_1 = sqrt(2^2 + 9^2) = sqrt85$

$theta_1 = tan ^ (-1) (-9/2) = -tan *-1 -4.5 = -77.47 = 282.53^@, " IV Quadrant"$

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

$theta_2 = tan ^ (-1) (2/6) = 18.43^@$

$z_1 / z_2 = sqrt(85/40) (cos (282.53- 18.43) - i sin (282.53 - 18.43))$

$color(crimson)(=> 1.33 ( -0.1028 - i 0.9947)$

How do you divide (2-9i)/(6+2i) (2-9i)/(6+2i) in trigonometric form?