Find the cube roots of 27(cos 279° + i sin 279°).

Solution:

Let Z = 27(cos 279° + i sin 279°)

Taking cube root on both sides

Z^1/3 = (27)^1/3 [cos279° + i sin 279° ]^1/3

Using De- Moivers theorem

If z = r [cos θ + i sinθ] then zⁿ = rⁿ[cos nθ + i sin nθ]

z^1/3 = (3³)^1/3 [cos(279°/3)+ i sin (279°/3)]

z^1/3 = 3[cos 93°+ i sin 93°]

Find the cube roots of 27(cos 279° + i sin 279°).

Summary:

The cube roots of 27(cos 279° + i sin 279°) is 3[cos93°+ i sin93°]