Ta có: $z_1\overline{z_1}=|z_1|^2=1 \Rightarrow \overline{z_1}=\frac{1}{z_1}$
Tương tự: $\overline{z_2}=\frac{1}{z_2}$
Ta có: $\overline{A}=\frac{\overline{z_1}+\overline{z_2}}{1+\overline{z_1.z_2}}=\frac{\displaystyle \frac{1}{z_1}+\frac{1}{z_2}}{\displaystyle 1+\frac{1}{z_1}.\frac{1}{z_2}}=\frac{z_1+z_2}{1+z_1.z_2}=A$
Vậy $A$ là số thực.