Question

If $n$ is a positive integer, then ${(\sqrt{3}+i)}^n+{(\sqrt{3}-i)}^n$ is equal to

• A. $2^n\cos \frac{n\pi }{6}$
• B. $2^{n+1}\cos \frac{n\pi }{6}$
• C. $2^{n-1}\cos \frac{n\pi }{6}$
• D. None of these

Answer 1

The correct option is B $2^{n+1}\cos \frac{n\pi }{6}$

Let $\sqrt{3}=r\cos \theta$ and $1=r\sin \theta$ so that

$r^2=4$ and $\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow r=2,\theta =\frac{\pi }{6}$

$\therefore {(\sqrt{3}+i)}^n=2^n{(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^n=2^n(\cos \frac{n\pi }{6}+i\sin \frac{n\pi }{6})$ ...(1)

Similarly $\therefore {(\sqrt{3}-i)}^n=2^n(\cos \frac{n\pi }{6}-i\sin \frac{n\pi }{6})$ ...(2)

Adding (1) and (2), we obtain

${(\sqrt{3}+i)}^n+{(\sqrt{3}-i)}^n=2.2^n\cos \frac{n\pi }{6}=2^{n+1}\cos \frac{n\pi }{6}$