Complex exercise asked in Lebanese university Math group by Hassan Kansoh

Exercise:

Find ${{z}_{1}}\,\And \,\,{{z}_{2}}$ such that $\frac{1}{{{z}_{1}}+{{z}_{2}}}=\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}$ where ${{z}_{1}},{{z}_{2}}\in \mathbb{C}$

Solution:

$\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}=\frac{{{z}_{2}}+{{z}_{1}}}{{{z}_{1}}{{z}_{2}}}=\frac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}{{z}_{2}}}=\frac{1}{{{z}_{1}}+{{z}_{2}}}\Leftrightarrow {{\left( {{z}_{1}}+{{z}_{2}} \right)}^{2}}={{z}_{1}}{{z}_{2}}$

$\Leftrightarrow {{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{1}}{{z}_{2}}=0$

But ${{z}_{1}}^{3}-{{z}_{2}}^{3}=\left( {{z}_{1}}-{{z}_{2}} \right)\left( {{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{1}}{{z}_{2}} \right)=\left( {{z}_{1}}-{{z}_{2}} \right)\left( 0 \right)=0$ so ${{z}_{1}}^{3}={{z}_{2}}^{3}\Leftrightarrow {{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}^{3}}=1$

Put $Z=\frac{{{z}_{1}}}{{{z}_{2}}}\Leftrightarrow {{Z}^{3}}={{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}^{3}}=1$

Let $Z=r{{e}^{i\theta }}\Leftrightarrow {{Z}^{3}}={{r}^{3}}{{e}^{i3\theta }}={{e}^{i2\pi }}=1$ So $r=1\,\,\And \,\,\theta =\frac{2\pi }{3}+\frac{2k\pi }{3}\,\,,\,\,\,k=0,1,2$

So $Z={{e}^{i\left( \frac{2\pi }{3}+\frac{2k\pi }{3} \right)}}$ i.e ${{z}_{1}}={{z}_{2}}{{e}^{i\left( \frac{2\pi }{3}+\frac{2k\pi }{3} \right)}}$