How do you divide $( 3i-7) / ( -3 i -9 )$ in trigonometric form?

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Answer 1 · 2018-06-24T03:47:06

$color(blue)(=> 0.8028 (0.7475 - i 0.6643)$

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

$z_1 = -7 + i 3, z_2 = -9 - i 3$

$r_1 = sqrt(7^2 + 3^2) = sqrt58$

$theta_1 = tan ^ (-1) (3/-7) = -tan *-1 (-3/7) = -23.2 ^@ = 156.8^@, " II Quadrant"$

$r_2 = sqrt(9^2 + 3^2) = sqrt90$

$theta_2 = tan ^ (-3/ -9) = tan^-1 (1/3) = 18.43^@ 198.43^@, " III Quadrant"$

$z_1 / z_2 = sqrt(58/90) (cos (156.8- 198.43) + i sin (156.8 - 198.43))$

$color(blue)(=> 0.8028 ( 0.7475 - i 0.6643)$

How do you divide ( 3i-7) / ( -3 i -9 )( 3i-7) / ( -3 i -9 ) in trigonometric form?