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

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Answer 1 · 2018-06-24T05:35:10

#color(blue)(=> 0.527 ( -0.9487 + i 0.3112)#

#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 = 3 + i 3#

#r_1 = sqrt(-2^2 + -1^2) = sqrt5#

#theta_1 = tan ^ (-1) (-1/-2) = tan ^-1 (1/2) = 26.57 ^@ = 206.57^@, " III Quadrant"#

#r_2 = sqrt(3^2 + 3^2) = sqrt 18#

#theta_2 = tan ^ (3/ 3) = tan^-1 (1) = 45^@#

#z_1 / z_2 = sqrt(5/8) (cos (206.57- 45) + i sin (206.57 - 45))#

#color(blue)(=> 0.527 ( -0.9487 + i 0.3112)#