How do you multiply # (9-2i)(2+i) # in trigonometric form?

Question

1446 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T05:13:43

#color(crimson)((9 - 2 i)* (2 + i) = 18.44 * (0.9701 + i 0.2426)#

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

#z_1 = (9 - 2 i), z_2 = (2 + i)#

#r_1 = sqrt (9^2 + -23^2) = sqrt (85)#

#theta _1 = tan ^-1 (-2/9) = 347.47^@, " IV quadrant"#

#r_2 = sqrt (2^2 + 1^2) = sqrt (5)#

#theta _2 = tan ^-1 (1/2) = 26.57^@ , " I quadrant"#

#z_1 * z_2 = (sqrt 85 * sqrt 5) * (cos(347.47 + 26.57) + i (347.47 + 26.57))#

#color(crimson)((9 - 2 i)* (2 + i) = 18.44 * (0.9701 + i 0.2426)#