Prove if $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} $ is a real number If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $
i found one link that had a similar problem. 
Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$
 A: We have $$\frac{z-w}{1-zw}=\frac{(z-w)(1-\overline{zw})}{(1-zw)(1-\overline{zw})}$$
Note that $(1-zw)(1-\overline{zw})=1-(zw+\overline{zw})+|zw|^2$ is real, so we need only consider the numerator (as Berci hinted).
We calculate $(z-w)(1-\overline{zw})=z-w-|z|^2\overline{w}+|w|^2\overline{z}=(z+\overline{z})-(w+\overline{w})$, which is real.
A: HINT:
Let $z=\cos A+i\sin A, w=\cos B+i\sin B$
$\implies z\cdot w=\cos(A+B)+i\sin(A+B)$
Using $\cos C-\cos D=-2\sin\frac{C-D}2\sin\frac{C+D}2$
and  $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2$
$$\frac{z-w}{1-zw}=\frac{\cos A-\cos B+i(\sin A-\sin B)}{1-\{\cos(A+B)+i\sin(A+B)\}}$$
$$=\frac{-2\sin\frac{A-B}2\sin\frac{A+B}2+i2\sin\frac{A-B}2\cos\frac{A+B}2}{2\sin^2\frac{A+B}2-i2\sin\frac{A+B}2\cos\frac{A+B}2}$$
$$=\frac{2i\sin\frac{A-B}2\left(i\sin\frac{A+B}2+\cos\frac{A+B}2\right)}{-2i\sin\frac{A+B}2\left(i\sin\frac{A+B}2+\cos\frac{A+B}2\right)}$$
$$=\frac{\sin\frac{A-B}2}{-\sin\frac{A+B}2}$$
EDIT:  The changed question can be addressed in the same way using
$\cos C+\cos D=2\cos\frac{C-D}2\cos\frac{C+D}2$
and  $\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$
$$\frac{z+w}{1+zw}=\frac{\cos A+\cos B+i(\sin A+\sin B)}{1+\{\cos(A+B)+i\sin(A+B)\}}$$
$$=\frac{2\cos\frac{A+B}2\cos\frac{A-B}2+i2\sin\frac{A+B}2\cos\frac{A-B}2}{2\cos^2\frac{A+B}2+i2\sin\frac{A+B}2\cos\frac{A+B}2}$$
$$=\frac{2\cos\frac{A-B}2\left(\cos\frac{A+B}2+i\sin\frac{A+B}2\right)}{2\cos\frac{A+B}2\left(\cos\frac{A+B}2+i\sin\frac{A+B}2\right)}$$
$$=\frac{\cos\frac{A-B}2}{\cos\frac{A+B}2}$$
A: Hint: For complex numbers $a,b$ we have $a/b\in\Bbb R\iff a\bar b\in\Bbb R$.
A: We have $|\zeta|=1 \Leftrightarrow \bar{\zeta}=\frac{1}{\zeta}$.
$|z|=|w|=1 \Leftrightarrow \bar{z}=\frac{1}{z} \text{and } \bar{w}=\frac{1}{w} \Leftrightarrow \frac{\bar{z}+\bar{w}}{1+\bar{z}\bar{w}}=\frac{\frac{1}{z}+\frac{1}{w}}{1+\frac{1}{z} \frac{1}{w}} \Leftrightarrow \overline{ \left(\frac{z+w}{1+zw}\right)}=\frac{z+w}{1+zw} \Leftrightarrow \frac{z+w}{1+zw} \in \mathbb R$ 
A: Draw vectors on the complex plane from the origin to a point on the unit circle. Notice that the argument of the sum of two such vectors is the average of the arguments of the two vectors (observe that the latter is defined only modulo a multiple of $\pi$). This follows from the properties of a rhombus (=parallelogram with equal sides).
Let $\arg z=\alpha$ and $\arg w=\beta$. By the above observation 
$\arg(z+w)=(\alpha+\beta)/2\pmod\pi$. It is well known that $\arg zw=\alpha+\beta$. As $\arg1=0$ another application of the above observation gives that 
$\arg(1+zw)=(\alpha+\beta)/2\pmod \pi$. 
So $z+w$ and $1+zw$ share the same argument modulo an integer multiple of $\pi$. This means that the argument of their ratio is a multiple of $\pi$, and hence the ratio itself is real.
A: HINT:
As $z \bar z=|z|^2=1$ and similarly for $w,$
$$\frac{z+w}{1+zw}=\frac{\frac1{\bar z}+\frac1{\bar w}}{1+\frac1{\bar z}\frac1{\bar w}}=\frac{\bar w+\bar z}{1+\bar w\bar z}$$
$$\text{Using  }\frac ab=\frac cd=\frac{a+c}{b+d},$$
$$\frac{z+w}{1+zw}=\frac{\bar w+\bar z}{1+\bar w\bar z}=\frac{z+w+\bar w+\bar z}{1+zw+1+\bar w\bar z}=\frac{(z+\bar z)+(w+\bar w)}{2+(zw+\overline{wz})}$$ as $\bar z\cdot \bar w=\overline{z\cdot w}$
Observe that each pair within parentheses is real
