How do you multiply # (1+2i)(2-5i) # in trigonometric form?

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Answer 1 · 2018-06-25T05:34:03

#color(orange)((1 + 2 i)* (2 - 5 i) = 12.04 * (0.9965 -+ i 0.0832)#

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

#z_1 = (1 + 2 i), z_2 = (2 - 5 i)#

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

#theta _1 = tan ^-1 (2/1) = 63.43^@, " I quadrant"#

#r_2 = sqrt (2^2 = -5^2) = sqrt (29)#

#theta _2 = tan ^-1 (-5/2) = 291.8^@ , " IV quadrant"#

#z_1 * z_2 = (sqrt 5 * sqrt 29) * (cos(63.43 + 291.8) + i (63.43 + 291.8))#

#color(orange)((1 + 2 i)* (2 - 5 i) = 12.04 * (0.9965 -+ i 0.0832)#