Question

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

• A. $16$
• B. $-16$
• C. $16\omega$
• D. $16{\omega }^2$

Answer 1

The correct option is C

$16\omega$

Explanation for the correct answer:

Step 1 : Information required for the question

We know that the cube root of unity is given by ${\mathit{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}$.

Also, $\mathbf{1}\mathbf{+}\mathit{\omega }\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{0}$ and from this equation we have,

$1+{\omega }^2=-\omega \dots \left(1\right)$

Step 2 : Calculation of the given term

First, 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)