Nice exercise using complex numbers to find its sum

Exercise:

Show that, $\sum\limits_{n=0}^{49}{\cos \left( \frac{n\pi }{3} \right)}=\frac{3}{2}$ then Deduce the value of $\sum\limits_{n=1}^{49}{\sin \left( \frac{n\pi }{3} \right)}$

Solution: Let $S=\sum\limits_{n=0}^{49}{\cos \left( \frac{n\pi }{3} \right)=\cos }\left( 0 \right)+\cos \left( \frac{\pi }{3} \right)+...+\cos \left( \frac{49\pi }{50} \right)$

Let ${{S}_{*}}=\sin \left( 0 \right)+\sin \left( \frac{\pi }{3} \right)+....+\sin \left( \frac{49\pi }{3} \right)$

Put $L=S+i{{S}_{*}}=\sum\limits_{n=0}^{49}{\cos \left( \frac{n\pi }{3} \right)+i\sum\limits_{n=0}^{49}{\sin \left( \frac{n\pi }{3} \right)}}$ where $i\in \mathbb{C}$

So $L=1+\cos \left( \frac{\pi }{3} \right)+\cos \left( \frac{2\pi }{3} \right)+...+\cos \left( \frac{49\pi }{3} \right)+i0+i\sin \left( \frac{\pi }{3} \right)+i\sin \left( \frac{2\pi }{3} \right)+...+i\sin \left( \frac{49\pi }{3} \right)$

$=1+\left( \cos \left( \frac{\pi }{3} \right)+i\sin \left( \frac{\pi }{3} \right) \right)+\left( \cos \left( \frac{2\pi }{3} \right)+i\sin \left( \frac{2\pi }{3} \right) \right)+...+\left( \cos \left( \frac{49\pi }{3} \right)+i\sin \left( \frac{49\pi }{3} \right) \right)$

Remark that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ whenever $r=1$

So $L=1+{{e}^{i\left( \frac{\pi }{3} \right)}}+{{e}^{i\left( \frac{2\pi }{3} \right)}}+....+{{e}^{i\left( \frac{49\pi }{3} \right)}}=1+{{e}^{i\left( \frac{\pi }{3} \right)}}+{{\left( {{e}^{i\frac{\pi }{3}}} \right)}^{2}}+....+{{\left( {{e}^{i\frac{\pi }{3}}} \right)}^{49}}$

Notice that $L$ represent a Geometric series of $50$ terms of first term is 1 and ratio is ${{e}^{i\frac{\pi }{3}}}$

So $L={{t}_{0}}\times \frac{1-{{r}^{50}}}{1-r}=1\times \frac{1-{{e}^{i\frac{50\pi }{3}}}}{1-{{e}^{i\frac{\pi }{3}}}}$

But \[{{e}^{i\frac{50\pi }{3}}}={{e}^{i\frac{\left( 48+2 \right)\pi }{3}}}={{e}^{i\frac{2\pi }{3}}}=\frac{-1}{2}+i\frac{\sqrt{3}}{2}\] And ${{e}^{i\frac{\pi }{3}}}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$

So $L=\frac{1+\frac{1}{2}-\frac{i\sqrt{3}}{2}}{1-\frac{1}{2}-i\frac{\sqrt{3}}{2}}=\frac{3-i\sqrt{3}}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}=\frac{3+3i\sqrt{3}-i\sqrt{3}-3{{i}^{2}}}{4}=\frac{6+2i\sqrt{3}}{4}=\frac{3}{2}+i\frac{\sqrt{3}}{2}$

Thus $\sum\limits_{n=1}^{49}{\sin \left( \frac{n\pi }{3} \right)=\frac{\sqrt{3}}{2}}$