Explanation:
B We have, \((1+\sqrt{3} \mathrm{i})^{\mathrm{n} / 2}\)
Now, \((1+\sqrt{3} i)^{n / 2}=\left[2\left\{\frac{1}{2}+\frac{i \sqrt{3}}{2}\right\}\right]^{n / 2}\)
\(=\left(-2 \omega^2\right)^{\mathrm{n} / 2}=(-2)^{\mathrm{n} / 2} \omega^{\mathrm{n}}\)
clearly, \((-2)^{\mathrm{n} / 2} \omega^{\mathrm{n}}\) is a positive real number
When, if \(\mathrm{n}=6,12,24,36\).
So, the smallest value of \(n=6\)