Complex Exercise asked by high school student in teachers' group

Exercise:

Solve in $\mathbb{C},$ $1+z+{{z}^{2}}+\bar{z}+{{\bar{z}}^{2}}=0$ such that ${{z}^{5}}=1\,\,\And \,z\ne 1$ where $ z\in \mathbb{C}$

Solution: Let $z={{e}^{i\theta }}$ then $\bar{z}={{e}^{-i\theta }}$ and by De Moiver’s ${{z}^{n}}={{e}^{in\theta }}$

So $1+{{e}^{i\theta }}+{{e}^{2i\theta }}+{{e}^{-i\theta }}+{{e}^{-i2\theta }}=0$ multiply by $\frac{1}{2}$ to get

$\frac{1}{2}+\frac{{{e}^{i\theta }}+{{e}^{-i\theta }}}{2}+\frac{{{e}^{i2\theta }}+{{e}^{-i2\theta }}}{2}=0$ but $\cos x=\frac{{{e}^{ix}}+{{e}^{-ix}}}{2}$

$\Rightarrow \frac{1}{2}+\cos \theta +\cos 2\theta =0$ but $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta -1$

$\Rightarrow 2{{\cos }^{2}}\theta +\cos \theta -\frac{1}{2}=0$ put $t=\cos \theta $

$\Rightarrow 2{{t}^{2}}+t-\frac{1}{2}=0\Rightarrow {{t}^{2}}+\frac{1}{2}t-\frac{1}{4}={{t}^{2}}+\frac{1}{2}t+\frac{1}{16}-\frac{1}{16}-\frac{1}{4}=0$

$\Rightarrow {{\left( t+\frac{1}{4} \right)}^{2}}=\frac{5}{16}\Leftrightarrow t=\frac{-1\pm \sqrt{5}}{4}$

If $t=\frac{-1+\sqrt{5}}{4}\Rightarrow \theta =\frac{2\pi }{5}+2k\pi $

If $t=\frac{-1-\sqrt{5}}{4}\Rightarrow \theta =\frac{4\pi }{5}+2k\pi $

Hence ${{z}_{1}}={{e}^{i\left( 2\pi /5 \right)}}\,\,\,\And \,\,\,\,{{z}_{2}}={{e}^{i\left( 4\pi /5 \right)}}$