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Question
Convert the complex numbers in polar form and also in exponential form.
`(-3)/2 + (3sqrt(3))/2"i"`
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Solution
Let z = `(-3)/2 + (3sqrt(3))/2"i"`
∴ a = `(-3)/2`, b = `(3sqrt(3))/2`, a < 0, b > 0
∴ r = `sqrt("a"^2 + "b"^2)`
= `sqrt(((-3)/2)^2 + ((3sqrt(3))/2)^2`
= `sqrt(9/4 + 27/4)`
= 3
Here `((-3)/2, (3sqrt(3))/2)` lies in 2nd quadrant
θ = amp (z) = `tan^-1 ("b"/"a") + pi`
= `tan^-1 (((3sqrt(2))/2)/((-3)/2)) + pi`
= `tan^-1(-sqrt(3)) + pi`
= `pi - pi/3`
= `(2pi)/3`
∴ θ = 120° = `(2pi)/3`
∴ The polar form of z = r(cos θ + i sin θ)
= 3(cos 120° + i sin 120°)
= `3(cos (2pi)/3 + "i" sin (2pi)/3)`
∴ The exponential form of z = reiθ = `3"e"^((2pi)/3"i")`
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