Question

If $n$ is an integer and
${(1+i\sqrt{3})}^n+{(1-i\sqrt{3})}^n=2^{n+1}\cos \theta ,$ then $\theta$ is equal to

• A. $\frac{n\pi }{3}$
• B. $\frac{n\pi }{2}$
• C. $\frac{n\pi }{4}$
• D. $\frac{n\pi }{6}$

Answer 1

The correct option is A $\frac{n\pi }{3}$

$2^n{(\frac{1}{2}+i\frac{\sqrt{3}}{2})}^n+2^n{(\frac{1}{2}-i\frac{\sqrt{3}}{2})}^n$
$=2^n{(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})}^n+2^n{(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})}^n$
Using De-Moivre's Theorem, we have
$=2^n(\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}+\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3})$
$=2^{n+1}\cos \frac{n\pi }{3}$
$\therefore \theta =\frac{n\pi }{3}$