Proving a Complex Number is Real I have the following question. 
Let $z_1$ and $z_2$ be complex numbers. 
Assumptions: 


*

*$|z_1|=|z_2|=1$

*$ z_1z_2 \neq -1$ 
What I have to prove is that: 
$$\frac{z_1+z_2}{1+z_1z_2}$$
is real.
My thoughts:
First, I multiplied the numerator and denominator by the conjugate of the denominator. Therefore, the denominator is real for sure, because of the proof I know that:
$$z\bar z = |z|^2 $$ 
Therefore it is real, and I can ignore the denominator.
The result is now:
$$(z_1 + z_2)(1 + \overline{z_1}\,\overline{z_2})$$
Another thought I had is to use trigonometric identities, but that do very well either.
Any help is appreciated.
 A: Hint: continue the expansion
$$
(z_1+z_2)(1+\overline{z_1}\,\overline{z_2})=
z_1+z_2+z_1\overline{z_1}\,\overline{z_2}+\overline{z_1}\,z_2\overline{z_2}
$$
A: Let be $\displaystyle w=\frac{z_{1}+z_{2}}{1+z_{1}z_{2}}$ and $w=a+ib$
We will prove that $w-\bar{w}=0$.
$$w-\bar{w}=\frac{z_{1}+z_{2}}{1+z_{1}z_{2}}-\frac{\bar z_{1}+\bar z_{2}}{1+\bar z_{1}\bar z_{2}}=\frac{z_{1}+\bar z_{2}+z_{2}+\bar z_{1}-\bar z_{1}-\bar z_{2}-z_{1}- z_{2}}{\left|1+z_{1}z_{2}\right|^{2}}=0$$
$$a+ib-a+ib=0 \Leftrightarrow b=0.$$ So $w$ real.
A: Since $z_1$ and $z_2$ lie on the unit circle they can be written as
$$
z_1=e^{i\alpha}, z_2=e^{i\beta}
$$
where $\alpha$ and $\beta$ are real. So
$$
\frac{z_1+z_2}{1+z_1 z_2}=\frac{(z_1+z_2)(1+\bar z_1\bar z_2)}{|1+z_1 z_2|^2}=\frac{(e^{i\alpha}+e^{i\beta})(1+e^{-i(\alpha+\beta)})}{|1+z_1 z_2|^2}=
$$
$$
\frac{e^{i\alpha}+e^{i\beta}+e^{-i\alpha}+e^{-i\beta}}{|1+z_1 z_2|^2}=\frac{2(\cos\alpha+\cos\beta)}{|1+z_1 z_2|^2} \in \mathbb R
$$
A: Hint: $$\text{Im}(z) = \frac{1}{2i}(z - \bar{z})$$
If you can prove that the imaginary part of a number is $0$, then it is real.
