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Question: Answered & Verified by Expert
If $\omega$ is the complex cube root of unity, the
$$
\left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2=
$$
MathematicsComplex NumberMHT CETMHT CET 2021 (22 Sep Shift 2)
Options:
  • A -1
  • B 0
  • C 4
  • D -4
Solution:
2378 Upvotes Verified Answer
The correct answer is: -4
$\begin{aligned} & \left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2 \\ & =\left(3+3 \omega+3 \omega^2+2 \omega\right)^2+\left(3+3 \omega+3 \omega^2+2 \omega^2\right)^2 \\ & =\left[3\left(1+\omega+\omega^2\right)+2 \omega\right]^2+\left[3\left(1+\omega+\omega^2\right)+2 \omega^2\right]^2 \\ & =(3(0)+2 \omega)^2+\left[3(0)+2 \omega^2\right]^2=4 \omega^2+4 \omega^4 \\ & =4 \omega^2\left(1+\omega^2\right)=4 \omega^2(-\omega) \\ & =-4 \omega^2=-4\end{aligned}$

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