Question

From the complex roots of unity in exponential form

Answer 1

Suppose let us assume that the cube root of 1 is z i.e., ∛1 = z.

Then, cubing both sides we get, z3 = 1

or, z3 - 1 = 0

or, (z - 1)(z2 + z + 1) = 0

Therefore, either z - 1 = 0 i.e., z = 1 or, z2 + z + 1 = 0

Therefore, z = −1±12−4⋅1⋅1√2⋅1 = −1±−3√2 = -12 ± i√32

Therefore, the three cube roots of unity are

1, -12 + i√32 and -12 - i√32

among them 1 is real number and the other two are conjugate complex numbers and they are also known as imaginary cube roots of unity.