What is $${\left[ {\frac{{\sin \frac{\pi }{6} + i\left( {1 - \cos \frac{\pi }{6}} \right)}}{{\sin \frac{\pi }{6} - i\left( {1 - \cos \frac{\pi }{6}} \right)}}} \right]^3}$$    where $$i = \sqrt { - 1} ,$$  equal to ?

Solution :
$$\eqalign{ & {\left[ {\frac{{\sin \frac{\pi }{6} + i\left( {1 - \cos \frac{\pi }{6}} \right)}}{{\sin \frac{\pi }{6} - i\left( {1 - \cos \frac{\pi }{6}} \right)}}} \right]^3} \cr & = {\left[ {\frac{{2\sin \frac{\pi }{{12}}\cos \frac{\pi }{{12}} + i\left( {2{{\sin }^2}\frac{\pi }{{12}}} \right)}}{{2\sin \frac{\pi }{{12}}\cos \frac{\pi }{{12}} - i\left( {2{{\sin }^2}\frac{\pi }{{12}}} \right)}}} \right]^3} \cr & = {\left[ {\frac{{\cos \frac{\pi }{{12}} + i\sin \frac{\pi }{{12}}}}{{\cos \frac{\pi }{{12}} - i\sin \frac{\pi }{{12}}}}} \right]^3} \cr & = {\left( {\frac{{{e^{i\frac{\pi }{{12}}}}}}{{{e^{ - i\frac{\pi }{12}}}}}} \right)^3} \cr & = {\left( {{e^{i\frac{\pi }{6}}}} \right)^3} \cr & = {e^{i \times 3 \times \frac{\pi }{6}}} \cr & = {e^{i\frac{\pi }{2}}} \cr & = \cos \frac{\pi }{2} + i\sin \frac{\pi }{2} \cr & = i \cr} $$