Question about proving that a fraction of complex numbers is real

Question

Let $z_1 , z_2$ be two complex numbers such that $|z_1|=|z_2| =1$ and $z_1z_2 \neq -1$. Prove that $\frac {z_1+z_2}{1+z_1z_2}$ is real.

I really don't know what to do with these types of problems. How can I show that some number is real? What conditions must this fraction satisfy to be able to say that it is real?

Answer 1

Hint Try showing that this complex number (call it $w$) satisfies $w=\bar{w}$. Also use the facts that $z_1\bar{z_1}=1$ and $z_2\bar{z_2}=1$.

Answer 2

Yet another way: $\displaystyle\;\frac {z_1+z_2}{1+z_1z_2} \cdot \color{red}{\frac{1+ \bar z_1 \bar z_2}{1+ \bar z_1 \bar z_2}} = \frac{z_1+\overbrace{|z_1|^2}^{=1}\bar z_2+z_2+\bar z_1 \overbrace{|z_2|^2}^{=1}}{|1+z_1z_2|^2} = \frac{2 \operatorname{Re}(z_1+z_2)}{|1+z_1z_2|^2}\,$.