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

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Answer 1 · 2018-06-25T04:18:07

#color(maroon)((3 - i)/ (4 + 2 i) = 1/2 * (-1 + i)#

#z-1 / z_2 = (r_1 * r_2) * ((cos (theta_1 - theta_2) + i sin (theta_1 - theta_2))#

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

#r_1 = sqrt(-3^2 + 1^2) = sqrt10#

#theta _ 1 = tan ^ -1 (1/-3) = -18.43 = 161.57^@, " II quadrant"#

#r_2 = sqrt(4^2 + 2^2) = sqrt20#

#theta _ 2 = tan ^ -1 (2/4) = 26.57^@, " I quadrant"#

#z_1 / z_2 = sqrt(10/20) * (cos (161.57 - 26.57) + i (161.57 - 26.57))#

#(3 - i)/ (4 + 2 i) = 1/sqrt2 * (-1/sqrt2 + i 1/sqrt2) = 1/2 * (-1 + i)#