How do you multiply # (-10-2i)(3-i) # in trigonometric form?

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Answer 1 · 2018-06-25T05:02:27

#color(magenta)((10 - 2 i)* (3 - i) = 32.25 * (-0.9923 + i 0.1239)#

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

#z_1 = (-10 - 2 i), z_2 = (3 - i)#

#r_1 = sqrt (-10^2 + -2^2) = sqrt (104)#

#theta _1 = tan ^-1 (-2/-10) + 191.31^@, " III quadrant"#

#r_2 = sqrt (3^2 = -1^2) = sqrt (10)#

#theta _2 = tan ^-1 (-1/3) = -18.43^@ = 341.57^@, " IV quadrant"#

#z_1 * z_2 = (sqrt104 * sqrt 10) * (cos(191.31 + 341.57) + i (191.31 + 341.57))#

#color(magenta)((10 - 2 i)* (3 - i) = 32.25 * (-0.9923 + i 0.1239)#