How do you divide (8+9i)/(1-3i) 8+9i1−3i in trigonometric form?

1 Answer
sankarankalyanam
Jun 24, 2018

color(violet)(=> 3.9115 ( -0.499 + i 0.8665)⇒3.9115(−0.499+i0.8665)

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 = -8 + i 9, z_2 = 1 - i 3z1=−8+i9,z2=1−i3

r_1 = sqrt(8^2 + 9^2) = sqrt 153r1=√82+92=√153

theta_1 = tan ^ (-1) (9/8) = -tan *-1 (1.125) = 48.37 ^@θ1=tan−1(98)=−tan⋅−1(1.125)=48.37∘

r_2 = sqrt(1^2 + (-3)^2) = sqrt 10r2=√12+(−3)2=√10

theta_2 = tan ^ (-3/ 1) = tan^-1 (-3) = -71.57^@ = 288.43^@, " IV Quadrant"θ2=tan−31=tan−1(−3)=−71.57∘=288.43∘, IV Quadrant

z_1 / z_2 = sqrt(153/10) (cos (48.37- 288.43) + i sin (48.37 - 288.43))z1z2=√15310(cos(48.37−288.43)+isin(48.37−288.43))

color(violet)(=> 3.9115 ( -0.499 + i 0.8665)⇒3.9115(−0.499+i0.8665)

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