Question

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

• A. $6$
• B. $10$
• C. $9$
• D. $12$

Answer 1

Answer: (D)

$12$

$(\sqrt{3}-i{)}^n=2^n$
$(2.e^{i\frac{-\pi }{6}}{)}^n=2^n$
$2^n{e}^{-i\frac{n\pi }{6}}=2^n$
$2^n(e^{-i\frac{n\pi }{6}}-1)=0$
Since $2^n\ne 0$,
Hence
$e^{-i\frac{n\pi }{6}}-1=0$
$e^{-i\frac{n\pi }{6}}=1$
$e^{-i\frac{n\pi }{6}}=e^{i2k\pi }$ where $k=\pm 1,\pm 2,\pm \mathrm{3..}$
$-\frac{n\pi }{6}=2k\pi$
$n=-12k$
Hence n is a multiples of 12.