How do you divide # (-4+5i)/(6+i) # in trigonometric form?

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Answer 1 · 2018-06-15T04:00:43

#color(purple)(=> 1.05 (-0.482 + i 0.876)#

#z_1 / z_2 = r_1/ r_2 (cos (theta_1 + thet_2) + i sin (theta_1 + theta_2)#

#z_1 = (-4 + i5), z_2 = (6 + i)#

#=> r_1 = sqrt(4^2 + 5^2) = sqrt41 = 6.4#

#theta_1 = tan^(-1) (5/-4)= -51.34^@ = 128.26^@, " II Quadrant"#

#=> r_2 = sqrt(1^2 + 6^2) = sqrt37 = 6.08#

#theta_2 = tan^(-1) (1/6)= 9.46^@ #

#z_1 / z_2 = sqrt41/sqrt37 (cos (128.26 - 9.46) + i sin(128.26 - 9.46)0#

#color(purple)(=> 1.05 (-0.482 + i 0.876)#