Q. If $n$ is an integer which leaves remainder one when divided by three, then $\left( 1 + \sqrt{3}i \right)^{n} + \left( 1 - \sqrt{3}i \right)^{n}$ equals
A) $- 2^{n + 1}$
B) $2^{n + 1}$
C) $- ( - 2)^{n}$
D) $- 2^{n}$
Explanation
(c)
Now, $\left( 1 + \sqrt{3}i \right)^{n} + \left( 1 - \sqrt{3}i \right)^{n}$
$= \left\lbrack 2\left( \frac{1 + \sqrt{3}i}{2} \right) \right\rbrack^{n} + \left\lbrack 2\left( \frac{1 - \sqrt{3}i}{2} \right) \right\rbrack^{n}$
$= \left( - 2\omega^{2} \right)^{n} + ( - 2\omega)^{n}$
$= ( - 2)^{n}\lbrack\left( \omega^{2} \right)^{3r + 1} + (\omega)^{3r + 1}\rbrack$
$\lbrack\because n = 3r + 1,\ where\ r\ is\ an\ integer\rbrack$
$= ( - 2)^{n}\left( \omega^{2} + \omega \right) = - ( - 2)^{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.