Question

If $(\sqrt{3}-i{)}^n=2^n,n\in N$, then $n$ is a multiple of

• A. $6$
• B. $10$
• C. $9$
• D. $12$

Answer 1

The correct option is D $12$

$\sqrt{3}-i=2\left(\frac{\sqrt{3}-i}{2}\right)$
$=2.e^{-i\frac{\pi }{6}}$
Hence, ${\left(2.e^{-i\frac{\pi }{6}}\right)}^n=2^n$

$2^n{e}^{-i\frac{n\pi }{6}}=2^n.e^{i2k\pi }$, $k\in N$

$\because e^{i2k\pi }=1$

Hence, $-\frac{n\pi }{6}=2k\pi$
$-n=12k$
Now $\cos (-x)=\cos x$
Hence, $n=12k$
Where $k$ is a natural number.