Complex exercise asked in brilliant.org

Exercise:

Show that , $\frac{{{\left( 1+i \right)}^{103}}}{{{\left( 1-i \right)}^{101}}}=-2$ where $i\in \mathbb{C}\,$

Solution: we have ${{z}_{1}}=1+i=\frac{\sqrt{2}}{\sqrt{2}}\left( 1+i \right)=\sqrt{2}\left( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right)=\sqrt{2}\left( \cos \left( \frac{\pi }{4} \right)+i\sin \left( \frac{\pi }{4} \right) \right)=\sqrt{2}{{e}^{i\frac{\pi }{4}}}$

So $\frac{{{\left( 1+i \right)}^{103}}}{{{\left( 1-i \right)}^{101}}}=\frac{{{\left( 1+i \right)}^{101}}{{\left( 1+i \right)}^{2}}}{{{\left( 1-i \right)}^{101}}}={{\left( 1+i \right)}^{2}}{{\left( \frac{1+i}{1-i} \right)}^{101}}={{\left( \sqrt{2}{{e}^{i\frac{\pi }{4}}} \right)}^{2}}{{\left( \frac{1+i}{1-i} \right)}^{101}}=2{{e}^{i\frac{\pi }{2}}}{{\left( \frac{1+i}{1-i} \right)}^{101}}$

$=2i{{\left( \frac{1+i}{1-i} \right)}^{101}}=\left( 2i \right){{\left( \frac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)} \right)}^{101}}=2i{{\left( \frac{1+{{i}^{2}}+2i}{1+i-i-{{i}^{2}}} \right)}^{101}}=2i{{\left( \frac{2i}{2} \right)}^{101}}=2{{i}^{102}}$

$=2{{\left( -1 \right)}^{51}}=-2$