[tex]|z_1|=|z_2|;\quad z_1\cdot z_2 \neq -1 \\ \\ |z_1|=1\Rightarrow \sqrt{z_1\cdot \overline{z_1}} = 1 \Rightarrow z_1\cdot \overline{z_1}=1,\quad z_2\cdot \overline{z_2} = 1[/tex]
[tex]Z = \dfrac{z_1+z_2}{1+z_1z_2}= \dfrac{\dfrac{z_1\cdot \overline{z_1}\cdot \overline{z_2}}{\overline{z_1}\cdot \overline{z_2}}+\dfrac{z_2\cdot \overline{z_1}\cdot \overline{z_2}}{\overline{z_1}\cdot \overline{z_2}}}{\dfrac{\overline{z_1}\cdot \overline{z_2}+(\overline{z_1}\cdot \overline{z_2})(z_1\cdot z_2)}{\overline{z_1}\cdot \overline{z_2}}}=[/tex]
[tex]=\dfrac{z_1\cdot \overline{z_1}\cdot \overline{z_2}+z_2\cdot \overline{z_2}\cdot \overline{z_1}}{\overline{z_1}\cdot \overline{z_2}+z_1\cdot \overline{z_1}+z_2\cdot \overline{z_2}} = \\ \\ = \dfrac{\overline{z_1}+\overline{z_2}}{1+\overline{z_1}\cdot \overline{z_2}} = \dfrac{\overline{z_1+z_2}}{1+\overline{z_1\cdot z_2}}=\dfrac{\overline{z_1+z_2}}{\overline{1+z_1\cdot z_2}} =\overline{Z} \\ \\ \\ Z = \overline{Z} \Rightarrow Z \in \mathbb{R}[/tex]
[tex]\text{Deoarece:}\\\\Z = a+bi,\quad \overline{Z} = a-bi \\ \\Z = \overline{Z} \Rightarrow a+bi = a-bi \Rightarrow b = 0 \Rightarrow \text{Im}(Z) = 0 \Rightarrow Z \in \mathbb{R}[/tex]
