2013 MJC P1 Q11

Show that \(\frac{1+\mathrm{i}\tan \theta }{1-\mathrm{i}\tan \theta }={{\mathrm{e}}^{\mathrm{i}\left( 2\theta \right)}}\) where \(-\frac{\pi }{2}<\theta <\frac{\pi }{2}\).

Hence find the value of \({{\left( \frac{1-\mathrm{i}\sqrt{3}}{1+\mathrm{i}\sqrt{3}} \right)}^{11}}\), giving your answer in the form \(r{{\mathrm{e}}^{\mathrm{i}\alpha }}\) where \(r>0\) and \(-\pi <\alpha \le \pi \).[5]