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MHT CET · Maths · Complex Number

If \(\mathrm{w}=\frac{-1+\mathrm{i} \sqrt{3}}{2}\), where \(\mathrm{i}=\sqrt{-1}\), then the value of \(\left(3+w+3 w^2\right)^4\) is

  1. A 16
  2. B -16
  3. C 16 w
  4. D \(16 w^2\)
Verified Solution

Answer & Solution

Correct Answer

(C) 16 w

Step-by-step Solution

Detailed explanation

\(\omega\) is a complex cube root of unity
\(\therefore \omega=1 ...(i)\)
\( \therefore 1+\omega+\omega^2=0, \)
\( \therefore \left(3+\omega+3 \omega^2\right)^4 ...(ii)\)
\( =(3+\omega+3(-1-\omega))^4 \)
\( =(3+\omega-3-3 \omega)^4 \)
\( =(-2 \omega)^4 \)
\( =16 \omega^4 \)
\( =16 \omega^3 \times \omega \)
\( =16 \omega\)
...[from (ii)]
...[from (i)]