How do you divide $( -2i+7) / ( -8i -3 )$ in trigonometric form?

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Answer 1 · 2018-06-15T04:22:33

$color(indigo)(=> 0.85 (-0.0804 + i 0.9968)$

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

$z_1 = 7 - i 20, z _2 = (-3 - i 8)$

$r_1 = sqrt(7^2 +2^2) = sqrt53$

$Theta_1 = arctan (-2/7) = 344.05^@, " IV Quadrant"$

$r_2 = sqrt(8^2 +3^2) = sqrt73$

$Theta_2 = arctan (-8/ -3) = 249.44^@, " III Quadrant"$

$z_1 / z_2 = (sqrt53 / sqrt73) (cos (344.05 - 249.44) + i sin(344.05 - 249.44))$

$color(indigo)(=> 0.85 (-0.0804 + i 0.9968)$

How do you divide ( -2i+7) / ( -8i -3 )( -2i+7) / ( -8i -3 ) in trigonometric form?