Q. If $a = \cos{\frac{4\pi}{3} + i\sin{\frac{4\pi}{3},}}$ then the value of $\left( \frac{1 + a}{2} \right)^{3n}$ is
A) $( - 1)^{n}$
B) $\frac{( - 1)^{n}}{2^{3n}}$
C) $\frac{1}{2^{3n}}$
D) $( - 1)^{n} + 1$
Explanation
(b)
$\frac{1 + a}{2} = \frac{1}{2}\left( 1 + \cos{\frac{4\pi}{3} + i\sin\frac{4\pi}{3}} \right)$
$= \frac{1}{2}2\cos{\frac{2\pi}{3}\left( \cos{\frac{2\pi}{3} + i\sin\frac{2\pi}{3}} \right)}$
$= - \frac{1}{2}\left( \cos{\frac{2\pi}{3} + i\sin\frac{2\pi}{3}} \right)$
$\therefore\ \ \left( \frac{1 + a}{2} \right)^{3n} = \left( \frac{- 1}{2} \right)^{3n}\left( \cos{\frac{2\pi}{3} + i\sin\frac{2\pi}{3}} \right)^{3n} = \frac{( - 1)^{n}}{2^{3n}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.