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$If\omega =\frac{-1 + \sqrt{3i}}{2}$ then $(3+\omega+\omega^{2})^4$ is

Answer:

Explanation:

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 )