11. (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
$$
\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}} \neq\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}} .
$$
11. Let $f_{\alpha}(z)=(z+\alpha)^{2}+\alpha \bar{z}$. Then
$$
\begin{array}{l}
f_{\alpha}\left(z_{1}\right)-f_{\alpha}\left(z_{2}\right)=\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}}-\left(z_{2}+\alpha\right)^{2}-\alpha \overline{z_{2}} \\
\quad=\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\left(\overline{z_{1}}-\overline{z_{2}}\right) .
\end{array}
$$

Assume 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
$$
\begin{array}{l}
\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\left(\overline{z_{1}}-\overline{z_{2}}\right)=0 \\
\Rightarrow \left|z_{1}+z_{2}+2 \alpha\right| \cdot \left|z_{1}-z_{2}\right| \leqslant|\alpha| \cdot \left|\overline{z_{1}}-\overline{z_{2}}\right| \\
\Rightarrow \left|z_{1}+z_{2}+2 \alpha\right| \cdot \left|z_{1}-z_{2}\right| \leqslant|\alpha| \cdot \left|z_{1}-z_{2}\right| \\
\Rightarrow \left|z_{1}+z_{2}+2 \alpha\right| \leqslant|\alpha| \\
\Rightarrow \left|z_{1}+z_{2}\right| \leqslant 2|\alpha|-2 \\
\Rightarrow|\alpha|<2 .
\end{array}
$$

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 $\left|-\frac{\alpha}{2} \pm \beta \mathrm{i}\right| \leqslant\left|-\frac{\alpha}{2}\right|+|\beta|<1$, so $\left|z_{1}\right|, \left|z_{2}\right|<1$. At this time, substituting
$$
\begin{array}{l}
z_{1}+z_{2}=-\alpha, z_{1}-z_{2}=2 \beta \mathrm{i}, \\
\overline{z_{1}}-\overline{z_{2}}=\overline{2 \beta \mathrm{i}}=-2 \beta \mathrm{i}
\end{array}
$$

into equation (1) yields
$$
\begin{array}{l}
f_{\alpha}\left(z_{1}\right)-f_{\alpha}\left(z_{2}\right)=\alpha \cdot 2 \beta \mathrm{i}+\alpha(-2 \beta \mathrm{i})=0 \\
\Rightarrow f_{\alpha}\left(z_{1}\right)=f_{\alpha}\left(z_{2}\right) .
\end{array}
$$

In summary, the values of $\alpha$ that meet the requirements are $\{\alpha|\alpha \in \mathbf{C},| \alpha \mid \geqslant 2\}$.