Example 10 Let the complex number $z$ satisfy the equation $|z-\mathrm{i}|=1$ and $z \neq 0, z \neq 2 \mathrm{i}$, and let the complex number $\omega$ be such that $\frac{\omega}{\omega-2 \mathrm{i}} \cdot \frac{z-2 \mathrm{i}}{z}$ is a real number. What is the set of points in the complex plane corresponding to $\omega$, and explain the reason.
Solve: Since $\frac{\omega}{\omega-2 \mathrm{i}} \cdot \frac{z-2 \mathrm{i}}{z}=\frac{(\omega-\mathrm{i})+\mathrm{i}}{(\omega-\mathrm{i})-\mathrm{i}} \cdot \frac{(z-\mathrm{i})-\mathrm{i}}{(z-\mathrm{i})+\mathrm{i}}$, let $A=\frac{\omega-\mathrm{i}}{-\mathrm{i}}, B=\frac{z-\mathrm{i}}{\mathrm{i}}$, then the above expression is $\frac{A-1}{A+1} \cdot \frac{B-1}{B+1}$. By $|z-\mathrm{i}|=1$ and $z \neq 0, z \neq 2 \mathrm{i}$, we know $|B|=1$ and $B \neq \pm 1 \Leftrightarrow \frac{B-1}{B+1}$ is a pure imaginary number. Therefore, by the given condition, $\frac{A-1}{A+1}$ is a pure imaginary number $\Leftrightarrow|A|=1$ and $A \neq \pm 1 \rightleftharpoons \omega \neq 0, \omega \neq 2 \mathrm{i}$, so $\omega$ corresponds to a set of points $z$ in the complex plane that form a circle with center $(0,1)$ and radius 1, but excluding the points $(0,0)$ and $(0,2)$.