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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!

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By $$ \overline x + x \, z _ 1 \, z _ 2 = z _ 1 + z _ 2 \tag { 0 } $$ we have $$ \overline { \overline x + x \, z _ 1 \, z _ 2 } = \overline { z _ 1 + z _ 2 } $$ $$ \therefore \quad x + \overline x \, \overline { z _ 1 } \, \overline { z _ 2 } = \overline { z _ 1 } + \overline { z _ 2 } \text {.} \tag { 1 } $$ Multiplying $ ( 0 ) $ by $ - \overline { z _ 1 } \, \overline { z _ 2 } $ and adding $ ( 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 $$ \big( 1 - | z _ 1 | ^ 2 \, | z _ 2 | ^ 2 \big) \, x = \overline { z _ 1 } \, \big( 1 - | z _ 2 | ^ 2 \big) + \overline { z _ 2 } \, \big( 1 - | z _ 1 | ^ 2 \big) \text {.} \tag { 2 } $$ Now since both $ | z _ 1 | $ and $ | z _ 2 | $ are real numbers less than $ 1 $, taking the modulus of the both sides of $ ( 2 ) $ and using triangle inequality, we get $$ \big( 1 - | z _ 1 | ^ 2 \, | z _ 2 | ^ 2 \big) \, | x | \le | z _ 1 | \, \big( 1 - | z _ 2 | ^ 2 \big) + | z _ 2 | \, \big( 1 - | z _ 1 | ^ 2 \big) \text {.} \tag { 3 } $$ It's now sufficient to prove $$ | z _ 1 | \, \big( 1 - | z _ 2 | ^ 2 \big) + | z _ 2 | \, \big( 1 - | z _ 1 | ^ 2 \big) < 1 - | z _ 1 | ^ 2 \, | z _ 2 | ^ 2 \tag { 4 } $$ since combining with $ ( 3 ) $ we can conclude that $ | x | < 1 $. It's a simple algebra exercise to show that $ ( 4 ) $ is equivalent to $$ ( 1 - | z _ 2 | ) \, ( 1 - | z _ 1 | ) \, ( 1 - | z _ 1 | \, | z _ 2 | ) > 0 \text {.} $$

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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 $\rho < 1$.

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