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