    Question

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

A

$16$

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B

$-16$

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C

$16\omega$

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D

$16{\omega }^{2}$

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Solution

## The correct option is C $16\omega$Explanation for the correct answer:Step 1 : Information required for the questionWe know that the cube root of unity is given by ${\mathbit{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}$.Also, $\mathbf{1}\mathbf{+}\mathbit{\omega }\mathbf{+}{\mathbit{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{0}$ and from this equation we have,$1+{\omega }^{2}=-\omega \dots \left(1\right)$Step 2 : Calculation of the given termFirst, simplify the expression ${\left(3+\omega +3{\omega }^{2}\right)}^{4}$,$\begin{array}{rcl}{\left(3+\omega +3{\omega }^{2}\right)}^{4}& =& {\left[3\left(1+{\omega }^{2}\right)+\omega \right]}^{4}\\ & =& {\left[-3\omega +\omega \right]}^{4}\left[\text{from}\left(1\right)\right]\\ & =& {\left[-2\omega \right]}^{4}\\ & =& 16{\omega }^{4}\\ & =& 16\omega ·{\omega }^{3}\left[\therefore {\omega }^{3}=1\right]\\ & =& 16\omega \end{array}$Hence, the correct option is (C)  Suggest Corrections  1      Similar questions  Explore more