How do you divide ( 2i -3) / (- 6 i -8 )2i−3−6i−8 in trigonometric form?

1 Answer
Kalyanam S.
Jul 26, 2018

color(crimson )(=> 0.12 - 0.34 i),⇒0.12−0.34i, IV QUADRANT

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 = -3 + 2 i, z_2 = -8 - 6 iz1=−3+2i,z2=−8−6i

r_1 = sqrt(-3^2 + 2^2)^2) = sqrt 13r1=√−32+222)=√13

theta_1 = tan ^ -1 (2 / -3) = 146.3099^@ = , " II Quadrant"θ1=tan−1(2−3)=146.3099∘=, II Quadrant

r_2 = sqrt(-8^2 + (-6)^2) = 10r2=√−82+(−6)2=10

theta_2 = tan ^-1 (-6/ -8) ~~ 216.8699^@, " III Quadrant"θ2=tan−1(−6−8)≈216.8699∘, III Quadrant

z_1 / z_2 = (sqrt(13) /10) * (cos (146.3099 - 216.8699) + i sin (146.3099 - 216.8699))z1z2=(√1310)⋅(cos(146.3099−216.8699)+isin(146.3099−216.8699))

color(crimson )(=> 0.12 - 0.34 i),⇒0.12−0.34i, IV QUADRANT

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