Question

Answer the following:

Convert the complex numbers in polar form and also in exponential form.

`(-3)/2 + (3sqrt(3))/2"i"`

Answer 1

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 2^nd 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 = re^iθ = `3"e"^((2pi)/3"i")`