How do you divide $( 2i -3) / (- 6 i -8 )$ in trigonometric form?

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Answer 1 · 2018-07-26T15:05:04

$color(crimson )(=> 0.12 - 0.34 i),$ IV QUADRANT

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

$z_1 = -3 + 2 i, z_2 = -8 - 6 i$

$r_1 = sqrt(-3^2 + 2^2)^2) = sqrt 13$

$theta_1 = tan ^ -1 (2 / -3) = 146.3099^@ = , " II Quadrant"$

$r_2 = sqrt(-8^2 + (-6)^2) = 10$

$theta_2 = tan ^-1 (-6/ -8) ~~ 216.8699^@, " III Quadrant"$

$z_1 / z_2 = (sqrt(13) /10) * (cos (146.3099 - 216.8699) + i sin (146.3099 - 216.8699))$

$color(crimson )(=> 0.12 - 0.34 i),$ IV QUADRANT

How do you divide ( 2i -3) / (- 6 i -8 )( 2i -3) / (- 6 i -8 ) in trigonometric form?