Find the roots of z^3=1 using De Moivre theorem

Answer 1

De Moivre's theorem states that for any complex number z = r(cos θ + i sin θ), and any positive integer n,

z^n = r^n (cos nθ + i sin nθ)

In this case, we have z^3 = 1, which means we can write:

z = cos(2πk/3) + i sin(2πk/3), where k = 0, 1, 2

Therefore, the roots of z^3 = 1 are:

z1 = cos(0) + i sin(0) = 1
z2 = cos(2π/3) + i sin(2π/3) = -1/2 + i(√3/2)
z3 = cos(4π/3) + i sin(4π/3) = -1/2 - i(√3/2)