How do you divide (1-6i) / (5-2i) 1−6i5−2i in trigonometric form?

1 Answer
sankarankalyanam
Jul 9, 2018

color(green)(=> 0.5861 - 0.9656 i)⇒0.5861−0.9656i

Explanation:

z_1 / z_2 = (r_1 / r_2) (cos (theta_1 - theta_2) + i sin (theta_1 - theta_2))z1z2=(r1r2)(cos(θ1−θ2)+isin(θ1−θ2))

z_1 = 1 - 6 i, z_2 = 5 - 2 iz1=1−6i,z2=5−2i

r_1 = sqrt(1^2 + -6^2) = sqrt 37r1=√12+−62=√37

theta_1 = tan ^ (-1) (-6/1) ~~ 279.46 ^@, " IV Quadrant"θ1=tan−1(−61)≈279.46∘, IV Quadrant

r_2 = sqrt(5^2 + (-2)^2) = sqrt 29r2=√52+(−2)2=√29

theta_2 = tan ^-1 (-2/ 5) ~~ 338.2^@, " IV Quadrant"θ2=tan−1(−25)≈338.2∘, IV Quadrant

z_1 / z_2 = sqrt(37/29) (cos (279.46- 338.2) + i sin (279.46- 338.2))z1z2=√3729(cos(279.46−338.2)+isin(279.46−338.2))

color(green)(=> 0.5861 - 0.9656 i)⇒0.5861−0.9656i

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1587 views around the world
You can reuse this answer
Creative Commons License