How do you divide #( i-2) / (i -6 )# in trigonometric form?

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Answer 1 · 2018-06-25T05:42:53

#color(maroon)((-2 + i) / (-6 + i) = 0.3676 ( 0.9957 - i 0.2942)#

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

#z_1 = -2 + i , z_2 = -6 + i #

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

#theta_1 = tan ^ (-1) (1/-2) = 153.43 ^@ " II Quadrant"#

#|r_2| = sqrt(-6^2 + (1)^2) = sqrt 37#

#theta_2 = tan ^-1 (1/ -6) = 170.54^@ , " II Quadrant"#

#z_1 / z_2 = |sqrt(5/37)| * (cos (153.43- 170.54) + i sin (143.43 - 170.54))#

#color(maroon)((-2 + i) / (-6 + i) = 0.3676 ( 0.9957 - i 0.2942)#