kingdukkwashere
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What is the polar form of Negative 9 minus 9 I StartRoot 3 EndRoot ?

9 (cosine (StartFraction pi over 3 EndFraction) + I sine (StartFraction pi Over 3 EndFraction) )
9 (cosine (StartFraction 4 pi over 3 EndFraction) + I sine (StartFraction 4 pi Over 3 EndFraction) )
18 (cosine (StartFraction pi over 3 EndFraction) + I sine (StartFraction pi Over 3 EndFraction) )
18 (cosine (StartFraction 4 pi over 3 EndFraction) + I sine (StartFraction 4 pi Over 3 EndFraction) )

Note: it is NOT 18(cos(pi/3)+isin(pi/3))

Sagot :

LammettHash

I think you mean the complex number

-9 - 9√3 i

This number has modulus

|-9 - 9√3 i| = √((-9)² + (-9√3)²) = √324 = 18

and argument θ such that

tan(θ) = (-9√3) / (-9) = √3

Since -9 - 9√3 i falls in the third quadrant of the complex plane, we expect θ to be between -π and -π/2 radians, so that

θ = arctan(√3) - π = π/3 - π = -2π/3

Then the polar form is

18 (cos(-2Ï€/3) + i sin(-2Ï€/3))

and -2Ï€/3 is the same angle as 2Ï€ - 2Ï€/3 = 4Ï€/3, so the correct choice is

18 (cos(4Ï€/3) + i sin(4Ï€/3))

Answer:

D.18 (cosine (StartFraction 4 pi over 3 EndFraction) + I sine (StartFraction 4 pi Over 3 EndFraction) )

Step-by-step explanation:

Got it right on Edge 2022

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