How do you multiply # (-1-5i)(3-6i) # in trigonometric form?

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Answer 1 · 2018-06-14T06:48:57

#color(crimson)((-1-i5) (3 - i 6) = (-33 - i 9)#

#z_1 = (-1 - i 5) #

#r_1 = sqrt(1^2 + 5^2) = sqrt26#

#theta_1 = arctan (-5/-1) = arctan 5 = 258.69^@, " III Quadrant"#

#z_2 = (3 - i 6)#

#r_2 = sqrt(3^2 + 6^2) = sqrt45#

#theta_2 = -6/3 = - 2 = 296.57^@, " IV Quadrant"#

#z_1 * z_2 = (r_1 * r_2) * (cos (theta-1 + theta_2) + i sin (theta_1 + theta_2))#

#z_1 * z_2 = (sqrt 26 * sqrt 45) * (cos (258.69 + 296.57) + i sin(258.69 + 296.57))#

#=> 34.21 (cos 555.26 + i sin 555.26) = 34.21(-0.9647 - i 0.2632)#

#color(crimson)((-1-i5) (3 - i 6) = (-33 - i 9)#