11. 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 $\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}} \neq\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}}$.
$$
\begin{array}{l}
\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}}=\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}} \Rightarrow\left(z_{1}+\alpha\right)^{2}-\left(z_{2}+\alpha\right)^{2}=\alpha\left(\overline{z_{2}}-\overline{z_{1}}\right) \\
\Rightarrow\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)=\alpha\left(\overline{z_{2}-z_{1}}\right) \Rightarrow\left|z_{1}+z_{2}+2 \alpha\right|\left|z_{1}-z_{2}\right|=|\alpha|\left|\overline{z_{2}-z_{1}}\right| . \\
\text { Since }\left|z_{1}-z_{2}\right|=\left|\overline{z_{2}-z_{1}}\right| \neq 0 \text {, then }|\alpha|=\left|z_{1}+z_{2}+2 \alpha\right| \geqslant 2|\alpha|-\left|z_{1}\right|-\left|z_{2}\right| \\
\Rightarrow|\alpha| \leqslant\left|z_{1}\right|+\left|z_{2}\right|<2 .
\end{array}
$$

On the other hand, if $|\alpha|<2$, take $z_{1}=-\frac{\alpha}{2}+\beta \mathrm{i}, z_{2}=-\frac{\alpha}{2}-\beta \mathrm{i}$, where $0<\beta<1-\left|\frac{\alpha}{2}\right|$, then $\left|z_{1}\right|=\left|z_{2}\right| \leqslant\left|\frac{\alpha}{2}\right|+|\beta|<1$.
At this point, $\left[\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}}\right]-\left[\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}}\right]=\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\left(\overline{z_{1}-z_{2}}\right)$ $=\alpha \cdot 2 \beta \mathrm{i}-\alpha \cdot 2 \beta \mathrm{i}=0 \Rightarrow\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}}=\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}}$.
In summary, all complex numbers $\alpha$ that meet the requirements are $\{\alpha \mid \alpha \in \mathbf{C}$, and $|\alpha| \geqslant 2\}$.