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

Hint: multiply numerator and denominator by $\bar{z}\bar{w}+1$ and recall that $z\bar{z}=w\bar{w}=1$; moreover $Z+\bar{Z}$ is real for every complex number $Z$.

$\dfrac{z+w}{zw+1}=\dfrac{(z+w)(\bar{z}\bar{w}+1)}{(zw+1)(\bar{z}\bar{w}+1)}=\dfrac{\bar{w}+\bar{z}+z+w}{2+zw+\bar{z}\bar{w}}=\dfrac{(z+\bar{z})+(w+\bar{w})}{(zw+\overline{zw})+2}$

