1) $If\omega =\frac{-1 + \sqrt{3i}}{2} $ then $ (3+\omega+\omega^{2})^4$ is A) 16 B) -16 C) $16\omega$ D) $16\omega^{2}$ Answer: Option CExplanation:Let $\omega =\frac{-1 + \sqrt{3i}}{2}$ Now, 1+ ω + ω2 = 0 and ω3= 1 $then (3+\omega+\omega^{2})^4$ = (3+ω+3ω2+2ω-2ω)4 =(3+3ω+3ω2-2ω)4 = [3(1+2+ω2)-2ω]4 = (0-2ω)4 = 16ω4 = 16ω (ω3 = 1 )
