11. (This question is worth 20 points) Determine all complex numbers $\alpha$ such that for any complex numbers $z_{1}, z_{2}\left(\left|z_{1}\right|,\left|z_{2}\right|<1, z_{1} \neq z_{2}\right)$, we have
$$
\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}} \neq\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}} .
$$
Let $f_{\alpha}(z)=(z+\alpha)^{2}+\alpha \bar{z}$. Then
$$
\begin{aligned}
f_{\alpha}\left(z_{1}\right) & -f_{\alpha}\left(z_{2}\right)=\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}}-\left(z_{2}+\alpha\right)^{2}-\alpha \overline{z_{2}} \\
= & \left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\left(\overline{z_{1}}-\overline{z_{2}}\right) .
\end{aligned}
$$

Suppose there exist complex numbers $z_{1}, z_{2}\left(\left|z_{1}\right|,\left|z_{2}\right|<1\right)$ such that $f_{\alpha}\left(z_{1}\right)=f_{\alpha}\left(z_{2}\right)$. Then
$$
\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\left(\overline{z_{1}}-\overline{z_{2}}\right)=0.
$$
Since $z_{1} \neq z_{2}$, we have
$$
z_{1}+z_{2}+2 \alpha+\alpha \frac{\overline{z_{1}}-\overline{z_{2}}}{z_{1}-z_{2}}=0.
$$
Let $z_{1}=x_{1}+y_{1} \mathrm{i}, z_{2}=x_{2}+y_{2} \mathrm{i}$, then
$$
\frac{\overline{z_{1}}-\overline{z_{2}}}{z_{1}-z_{2}}=\frac{x_{1}-x_{2}-(y_{1}-y_{2}) \mathrm{i}}{x_{1}-x_{2}+(y_{1}-y_{2}) \mathrm{i}}=\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}=1.
$$
Thus,
$$
z_{1}+z_{2}+2 \alpha+\alpha=0 \Rightarrow z_{1}+z_{2}+3 \alpha=0.
$$
Since $|z_{1}|,|z_{2}|<1$, we have $|z_{1}+z_{2}| \leq |z_{1}|+|z_{2}|<2$, so
$$
|3 \alpha| \leq 2 \Rightarrow |\alpha| \leq \frac{2}{3}.
$$
However, this contradicts the condition $|\alpha|<2$. Therefore, if $|\alpha| \geq 2$, there do not exist $z_{1}, z_{2}\left(\left|z_{1}\right|,\left|z_{2}\right|<1\right)$ such that $f_{\alpha}\left(z_{1}\right)=f_{\alpha}\left(z_{2}\right)$.

On the other hand, for any complex number $\alpha$ satisfying $|\alpha|<2$, let $z_{1}=-\frac{\alpha}{2}+\beta \mathrm{i}, z_{2}=-\frac{\alpha}{2}-\beta \mathrm{i}$, where $0<\beta<1-\frac{|\alpha|}{2}$. Then $z_{1} \neq z_{2}$, and $\left|-\frac{\alpha}{2} \pm \beta \mathrm{i}\right| \leq\left|-\frac{\alpha}{2}\right|+|\beta|<1$, so $\left|z_{1}\right|,\left|z_{2}\right|<1$. At this point, substituting
$$
z_{1}+z_{2}=-\alpha, \quad z_{1}-z_{2}=2 \beta \mathrm{i}, \quad \overline{z_{1}}-\overline{z_{2}}=\overline{2 \beta \mathrm{i}}=-2 \beta \mathrm{i}
$$
into (1) yields $f_{\alpha}\left(z_{1}\right)-f_{\alpha}\left(z_{2}\right)=\alpha \cdot 2 \beta \mathrm{i}+\alpha \cdot(-2 \beta \mathrm{i})=0$, i.e., $f_{\alpha}\left(z_{1}\right)=f_{\alpha}\left(z_{2}\right)$.
In conclusion, the values of $\alpha$ that meet the requirements are $\{\alpha|\alpha \in \mathbf{C},| \alpha \mid \geq 2\}$.
$\cdot 20$ points