Example 4 Determine all complex numbers $\alpha$ such that for any complex numbers $z_{1}, z_{2}\left(\left|z_{1}\right|<1,\left|z_{2}\right|<1, z_{1} \neq z_{2}\right)$, we have $\left(z_{1}+\alpha\right)^{2}+\alpha \overline{z_{1}} \neq\left(z_{2}+\alpha\right)^{2}+\alpha \overline{z_{2}}$.
【Analysis】If it is difficult to solve the problem directly, consider the opposite of the given conditions, which is equality, to derive useful information more easily. Therefore, consider the opposite of the problem.
Solution Let $f(z)=(z+\alpha)^{2}+\alpha \bar{z}$. Then
$$
\begin{array}{l}
f\left(z_{1}\right)-f\left(z_{2}\right) \\
=\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)+\alpha\left(\overline{z_{1}}-\overline{z_{2}}\right) .
\end{array}
$$

Assume there exists $\left|z_{1}\right|2|\alpha|-2 \\
\Rightarrow|\alpha|<2 .
\end{array}
$$

Next, we prove that $|\alpha|<2$ is also sufficient, i.e., there exist $\left|z_{1}\right|<1,\left|z_{2}\right|<1\left(z_{1} \neq z_{2}\right)$ such that
$$
f\left(z_{1}\right)=f\left(z_{2}\right),
$$

which means $\left(z_{1}+z_{2}+2 \alpha\right)\left(z_{1}-z_{2}\right)=-\alpha\left(\overline{z_{1}}-\overline{z_{2}}\right)$.
Suppose $z_{1} \backslash z_{2}$ are "conjugate," then $\overline{z_{1}}-\overline{z_{2}}$ differs from $z_{1}-z_{2}$ by a negative sign, and it suffices to have $z_{1}+z_{2}+2 \alpha=\alpha$.

Thus, we can 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 $\left|z_{1,2}\right| \leqslant \frac{|\alpha|}{2}+|\beta|<1$, satisfying:
$$
\begin{array}{l}
z_{1}+z_{2}=-\alpha, \\
z_{1}-z_{2}=2 \beta \mathrm{i}=-\left(\overline{z_{1}}-\overline{z_{2}}\right) .
\end{array}
$$

Therefore, $f\left(z_{1}\right)-f\left(z_{2}\right)=0$ meets the requirement.
In summary, the $\alpha$ that satisfies the requirement is
$$
\{\alpha|\alpha \in \mathbf{C},| \alpha \mid \geqslant 2\} .
$$