Vivekanand Vellanki
1
Let the fourth root be x = r\times(\cos(\alpha) + i \sin(\alpha)). Then, x^4 = -6 -6i\sqrt{3}. Substituting the value of x, we get (r\times(\cos(\alpha) + i\sin(\alpha)))^4 = -6 -6i\sqrt{3}. Using De Moivre's theorem, we get r^4\times(\cos (4\alpha) +i\sin(4\alpha)) = 12(-1/2 - \sqrt{3}/2) This gives, r^4=12; and \cos(4\alpha) = -1/2 and \sin(4\alpha) = -\sqrt{3}/2 \cos(240^{\circ}+360^{\circ}n) = -1/2 \text{ and } \sin(240^{\circ}+360^{\circ}n) = -\sqrt{3}/2. Hence, \alpha = 60^{\circ} + 90^{\circ}n; and $r=\sqrt[4]{12}$. We get distinct roots for n = 0, 1, 2 \text{ and } 3 . The fourth roots are $\sqrt[4]{12}(\cos(60^{\circ}) + i \sin(60^{\circ})), \sqrt[4]{12} (\cos(150^{\circ}) + i \sin(150^{\circ})), \sqrt[4]{12} (\cos(240^{\circ}) + i \sin(240^{\circ})) \text{ and } \sqrt[4]{12} (\cos(330^{\circ}) + i \sin(330^{\circ}))$.