How do you divide $(8+i)/(-2-9i)$ in trigonometric form?

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Answer 1 · 2018-07-26T14:29:06

$color(blue)(=> -0.2941 + 0.8235 i), II$ QUADRANT

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

$z_1 = 8 + i, z_2 = -2 - 9 i$

$r_1 = sqrt(8^2 + 1^2)^2) = sqrt 65$

$theta_1 = tan ^ (1)/ (8) = 7.125^@ = , " I Quadrant"$

$r_2 = sqrt(-2^2 + (-9)^2) = sqrt 85$

$theta_2 = tan ^-1 (-9/ -2) ~~ 257.4712^@, " III Quadrant"$

$z_1 / z_2 = sqrt(65 / 85) (cos (7.125 - 257.4712) + i sin (7.125 - 257.4712))$

$color(blue)(=> -0.2941 + 0.8235 i)$

How do you divide (8+i)/(-2-9i) (8+i)/(-2-9i) in trigonometric form?