Question

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

• A. $6$
• B. $10$
• C. $9$
• D. $12$

Answer 1

The correct option is D $12$

We have ${(\sqrt{3}-i)}^n=2^n,n\in Z$
$\Rightarrow {\left(\frac{\sqrt{3}-i}{2}\right)}^n=1$
$\Rightarrow i^n{(-\frac{1}{2}-\frac{\sqrt{3}-i}{2})}^n=1$
$\Rightarrow i^n{\omega }^{2n}=1$
$\Rightarrow i^n=1;{\omega }^{2n}=1$
$\Rightarrow$ $n$ is a multiple of $3$ and $4$ both.
$\Rightarrow$ $n$ is a multiple of $12$.