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

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Answer 1 · 2018-06-25T05:56:42

#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) = 26.57 ^@ " I Quadrant"#

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

#theta_2 = tan ^-1 (-1/ 6) = 350.54^@ , " IV Quadrant"#

#z_1 / z_2 = |sqrt(5/37)| * (cos (26.57 - 350.54) + i sin (26.57 - 350.54))#

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