z=1/2-i√3/2 , find z^3

Question

This is from the topic complex number

Answer 1

z=1/2-i√3/2 , find z^3

This complex number can be written as,

z = cos(-60) + i.sin(-60)

Using de Moivre’s formula, viz. (cos(θ) + i.sin(θ))^n = cos(n θ) + i.sin(n θ),

z^3 = (cos(-60) + i.sin(-60))^3 = cos(3*(-60)) + i.sin(3*(-60))

z^3 = cos(-180) + i.sin(-180)

z^3 = cos(360 – 180) + i.sin(360 – 180)

z^3 = cos(180) + i.sin(180)

z^3 = -1 + i.0

z^3 = -1