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

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Answer 1 · 2018-07-08T03:45:49

#color(maroon)(=> 1.06 + i 2.1196)#

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

#z_1 = 8 + i 3, z_2 = 2 - i 3#

#r_1 = sqrt(8^2 + 3^2) = sqrt 73#

#theta_1 = tan ^ (-1) (1/8) = 7.13 ^@, " I Quadrant"#

#r_2 = sqrt(2^2 + (-3)^2) = sqrt 13#

#theta_2 = tan ^-1 (-3/ 2) = 303.69^@, " IV Quadrant"#

#z_1 / z_2 = sqrt(73/13) (cos (7.13- 303.69) + i sin (7.13- 303.69))#

#color(maroon)(=> 1.06 + i 2.1196)#