The smallest positive integral value of $$n$$ for which $${\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}}$$  is real is

Solution :
$$\eqalign{ & {\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}} = {\left\{ {2\left( {\frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)} \right\}^{\frac{n}{2}}} \cr & {\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}} = {2^{\frac{n}{2}}}{\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right)^{\frac{n}{2}}} \cr & {\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}} = {2^{\frac{n}{2}}}{\left( {\cos \frac{{n\pi }}{3} + i\sin \frac{{n\pi }}{3}} \right)^{\frac{1}{2}}}. \cr} $$
Clearly, the least positive integral value of $$n$$ for which $$\cos\frac{{n\pi }}{3} + i\sin \frac{{n\pi }}{3}$$    is positive real is $$6.$$