Show that $x$ is within the unit disk if $\overline{x} + xz_1z_2 = z_1 + z_2$ and $|z_1|,|z_2| < 1$. Let $x \in \mathbb{C}$. Denote the complex conjugate of $x$ as $\overline{x}$. Fix $z_1, z_2 \in \mathbb{C}$ such that $|z_1| < 1$ and $|z_2| <1$. 
The following equation is satisfied:
$$\overline{x} + xz_1z_2 = z_1 + z_2$$
I want to show that $|x| < 1$. I tried a whole bunch of numerics and this seems to be true, but I have difficulty proving it. Any suggestions would be appreciated!
 A: The case $x = 0$ is uninteresting, so assume that $x \neq 0$.
Write $x = \rho e^{i \alpha}$ with $\rho > 0$ and put $w_j = e^{i \alpha} z_j$ for $j = 1,2$.
Multiplying the given equation by $e^{i\alpha}$, we get
$\rho (1 + w_1 w_2) = w_1 + w_2$, or
$$
\rho = \frac{w_1 + w_2}{1 + w_1 w_2}.
$$
We claim $0 < \rho < 1$ and note $\rho = f(w_1)$, where $f(z) = \dfrac{z + w_2}{1 + z w_2}$.
Recall that fractional linear transformations preserve generalized circles. 
The fractional linear transformation $f^{-1}(z) = \dfrac{z - w_2}{1 - w_2 z}$ fixes $\pm 1$ and sends the interval $[-1, 1]$ onto a circle segment through $\pm 1$ and $-w_2 = f^{-1}(0)$ inside the unit disk. Moreover, $f^{-1}(\mathbb{R} \setminus [-1,1])$ lies outside the unit disk.
As $\rho > 0$ by assumption and $w_1 = f^{-1}(\rho)$ lies inside the unit disk, we conclude $\rho < 1$.
A: By
$$ \overline x + x \, z _ 1 \, z _ 2 = z _ 1 + z _ 2 \tag 0 \label 0 $$
we have
$$ \overline { \overline x + x \, z _ 1 \, z _ 2 } =
\overline { z _ 1 + z _ 2 } $$
$$ \therefore \quad x + \overline x \, \overline { z _ 1 } \,
\overline { z _ 2 } = \overline { z _ 1 } + \overline { z _ 2 }
\text . \tag 1 \label 1 $$
Multiplying \eqref{0} by $ - \overline { z _ 1 } \, \overline { z _ 2 } $
and adding \eqref{1} we get
$$ \big( 1 - z _ 1 \, \overline { z _ 1 } \, z _ 2 \, \overline { z _ 2 }
\big) \, x = \overline { z _ 1 } \, \big( 1 - z _ 2 \, \overline { z _ 2 }
\big) + \overline { z _ 2 } \, \big( 1 - z _ 1 \, \overline { z _ 1 } \big) $$
which because of $ | z _ 1 | ^ 2 = z _ 1 \, \overline { z _ 1 } $ and
$ | z _ 2 | ^ 2 = z _ 2 \, \overline { z _ 2 } $ is equivalent to
$$ \big( 1 - | z _ 1 | ^ 2 \, | z _ 2 | ^ 2 \big) \, x
= \overline { z _ 1 } \, \big( 1 - | z _ 2 | ^ 2 \big) + \overline
{ z _ 2 } \, \big( 1 - | z _ 1 | ^ 2 \big) \text . \tag 2 \label 2 $$
Now since both $ | z _ 1 | $ and $ | z _ 2 | $ are real numbers less
than $ 1 $, taking the modulus of the both sides of \eqref{2}
and using triangle inequality, we get
$$ \big( 1 - | z _ 1 | ^ 2 \, | z _ 2 | ^ 2 \big) \, | x |
\le | z _ 1 | \, \big( 1 - | z _ 2 | ^ 2 \big) +
| z _ 2 | \, \big( 1 - | z _ 1 | ^ 2 \big) \text . \tag 3 \label 3 $$
It's now sufficient to prove
$$ | z _ 1 | \, \big( 1 - | z _ 2 | ^ 2 \big) +
| z _ 2 | \, \big( 1 - | z _ 1 | ^ 2 \big) <
1 - | z _ 1 | ^ 2 \, | z _ 2 | ^ 2 \tag 4 \label 4 $$
since combining with \eqref{3} we can conclude that $ | x | < 1 $.
It's a simple algebra exercise to show that \eqref{4} is equivalent to
$$ ( 1 - | z _ 2 | ) \, ( 1 - | z _ 1 | ) \,
( 1 - | z _ 1 |  \, | z _ 2 | ) > 0 \text . $$
