How do you divide #(-i+2) / (3i-4)# in trigonometric form?

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Answer 1 · 2018-06-25T06:08:35

#color(purple)((2 -i) / (-4 + 3i) = 2.236 ( 0.9839 - i 0.1788)#

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

#z_1 = 2 - i , z_2 = -4 + 3 i #

#|r_1| = sqrt(2^2 + -1^2) = sqrt 5#

#theta_1 = tan ^ (-1) (-1/2) = 333.43 ^@ " IV Quadrant"#

#|r_2| = sqrt(-4^2 + (3)^2) = 5#

#theta_2 = tan ^-1 (3/ -4) = 143.13^@ , " II Quadrant"#

#z_1 / z_2 = |(5/sqrt5)| * (cos (333.43 - 143.13) + i sin (333.43 - 143.13))#

#color(purple)((2 -i) / (-4 + 3i) = 2.236 ( 0.9839 - i 0.1788)#