If the cube roots of unity are 1, $$\omega \,,\,{\omega ^2}$$ then the roots of the equation $${(x - 1)^3}$$ + 8 = 0, are :

If $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$, then $$z$$ lies on :

If $${z_1}$$ and $${z_2}$$ are two non-zero complex numbers such that $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$, then arg $${z_1}$$ - arg $${z_2}$$ is equal to :

Let z and w be complex numbers such that $$\overline z + i\overline w = 0$$ and arg zw = $$\pi $$. Then arg z equals :