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
 A: Since $|z|=|w|=1$, 
$$z\bar z=w\bar w=1.$$
Letting $u$ be the given number, 
$$u=\frac{z+w}{zw+1}=\frac{(1/\bar z)+(1/\bar w)}{zw+1}=\frac{\bar w+\bar z}{\bar z\bar w(zw+1)}=\frac{\bar z+\bar w}{\bar z\bar w+1}=\bar u.$$
A: Hint: A complex number $z$ is real if and only if $z = \bar{z}$.
A: 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 $Z$.

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

A: Try ${ z+w \over zw+1} = { (z+w)(1+\bar{z}\bar{w}) \over |zw+1|^2} ={ z+w+\bar{w}+\bar{z} \over |zw+1|^2} = {2\over |zw+1|^2} \operatorname{re} (z+w)$.
A: Let $\displaystyle z=\cos2\alpha+i\sin2\alpha,w=\cos2\beta+i\sin2\beta$
$\implies z+w=\cos2\alpha+\cos2\beta+i\sin2\alpha+i\sin2\beta$
using Prosthaphaeresis Formulas,
$\displaystyle z+w=2\cos(\alpha-\beta)\{\cos(\alpha+\beta)+i\sin(\alpha+\beta)\}\  \ \  \ (1) $
$\displaystyle\implies zw=(\cos2\alpha+i\sin2\alpha)(\cos2\beta+i\sin2\beta)=\cos2(\alpha+\beta)+i\sin2(\alpha+\beta)$
$\displaystyle\implies zw+1=\cos2(\alpha+\beta)+i\sin2(\alpha+\beta)+1$
$\displaystyle=2\cos^2(\alpha+\beta)+2\sin(\alpha+\beta)\cos(\alpha+\beta)$
$\displaystyle\implies zw+1=2\cos(\alpha+\beta)\{\cos(\alpha+\beta)+i\sin(\alpha+\beta)\   \   \  \ (2)$
A: As $\displaystyle|z|=1, z\bar z=1$
Similarly, $\displaystyle w\bar w=1\implies \overline{zw}=\bar z\bar w=\frac1{zw}$
If $\displaystyle a=\frac{z+w}{zw+1},$
$\displaystyle \bar a=\frac{\bar z+\bar w}{\overline{zw+1}}=\frac{\bar z\bar w}{\overline{z w}+1}=\frac{\frac1z+\frac1w}{\frac1{zw}+1}=\frac{w+z}{1+zw}=a$
