Question 118, Determine all complex numbers $\alpha$ such that for any complex numbers $\mathrm{z}_{1} 、 \mathrm{z}_{2}\left(\left|\mathrm{z}_{1}\right| 、\left|\mathrm{z}_{2}\right|<1, \mathrm{z}_{1} \neq \mathrm{z}_{2}\right)$, we have $\left(\mathrm{z}_{1}+\alpha\right)^{2}+\alpha \overline{\mathrm{z}_{1}} \neq\left(\mathrm{z}_{2}+\alpha\right)^{2}+\alpha \overline{\mathrm{z}_{2}}$.
Question 118, Solution: We consider the necessary and sufficient condition for the existence of complex numbers $z_{1}, z_{2}, |z_{1}|, |z_{2}| < 1, z_{1} \neq z_{2}$, such that $(z_{1} + \alpha)^{2} + \alpha \overline{z_{1}} = (z_{2} + \alpha)^{2} + \alpha \overline{z_{2}}$. At this point,
$$
\begin{array}{l}
(z_{1} + \alpha)^{2} + \alpha \overline{z_{1}} = (z_{2} + \alpha)^{2} + \alpha \overline{z_{2}} \\
\Leftrightarrow \alpha (\overline{z_{1}} - \overline{z_{2}}) = (z_{2} - z_{1})(z_{1} + z_{2} + 2 \alpha) \\
\Leftrightarrow \alpha \left[\overline{z_{1} - z_{2}} + 2(z_{1} - z_{2})\right] = (z_{2} - z_{1})(z_{1} + z_{2}) \\
\Leftrightarrow \alpha \left(\frac{\overline{z_{1} - z_{2}}}{z_{2} - z_{1}} - 2\right) = z_{1} + z_{2}
\end{array}
$$

On one hand, since $|z_{1} + z_{2}| = |\alpha| \cdot \left|\frac{\overline{z_{1} - z_{2}}}{z_{2} - z_{1}} - 2\right| \geq |\alpha| \cdot \left(2 - \left|\frac{z_{1} - z_{2}}{z_{2} - z_{1}}\right|\right) = |\alpha|$, and $|z_{1} + z_{2}| \leq |z_{1}| + |z_{2}| < 2$, hence $|\alpha| < 2$.

On the other hand, if $|\alpha| < 2$, take a positive real number $\beta < 2 - |\alpha|$. Let $\left\{\begin{array}{l}z_{2} - z_{1} = \beta i \\ z_{1} + z_{2} = -\alpha\end{array}\right.$, i.e., $\left\{\begin{array}{l}z_{1} = \frac{-\alpha - \beta i}{2} \\ z_{2} = \frac{-\alpha + \beta i}{2}\end{array}\right.$, then $|z_{1}| = \left|\frac{-\alpha - \beta i}{2}\right| \leq \frac{|\alpha| + |\beta|}{2} < \frac{|\alpha| + (2 - |\alpha|)}{2} = 1$, and $|z_{2}| = \left|\frac{-\alpha + \beta i}{2}\right| \leq \frac{|\alpha| + |\beta|}{2} < \frac{|\alpha| + (2 - |\alpha|)}{2} = 1$, and $\alpha \left(\frac{\overline{z_{1} - z_{2}}}{z_{2} - z_{1}} - 2\right) = \alpha \left(\frac{\frac{-\beta i}{\beta i}}{\beta i} - 2\right) = -\alpha = z_{1} + z_{2}$.

Combining the above two aspects, when and only when $|\alpha| < 2$, there exist complex numbers $z_{1}, z_{2}, |z_{1}|, |z_{2}| < 1, z_{1} \neq z_{2}$, such that $(z_{1} + \alpha)^{2} + \alpha \overline{z_{1}} = (z_{2} + \alpha)^{2} + \alpha \overline{z_{2}}$. Therefore, the values of $\alpha$ that satisfy for any complex numbers $z_{1}, z_{2} (|z_{1}|, |z_{2}| < 1, z_{1} \neq z_{2})$, $(z_{1} + \alpha)^{2} + \alpha \overline{z_{1}} \neq (z_{2} + \alpha)^{2} + \alpha \overline{z_{2}}$ are $\{\alpha | \alpha \in \mathbb{C}, |\alpha| \geq 2\}$.