Complex numbers - proof Let z and w be complex numbers such that $|z| = |w| = 1$ and $zw \neq -1$. Prove that $\frac{z + w}{zw + 1}$ is a real number.
I let z = a + bi and w = c+ di so we have that $\sqrt{a^2+b^2} = \sqrt{c^2+d^2} = 1$ and $a^2+b^2 = c^2 + d^2 = 1$. I plugged it into the equation but I didn't really get anything worth noting. Does anyone know how to solve this? Thanks

Since $|z|=|w|=1$, 
$$z\bar z=w\bar w=1.$$
Letting $u$ be the given number, 
$$u=\frac{z+w}{zw+1}=\frac{(1/\bar z)+(1/\bar w)}{zw+1}=\frac{\bar w+\bar z}{\bar z\bar w(zw+1)}=\frac{\bar z+\bar w}{\bar z\bar w+1}=\bar u.$$
