Example 7 (Mathematical Olympiad Training Question from "Intermediate Mathematics" Issue 6, 2003) Let the complex number $z$ satisfy the condition $|z-\mathrm{i}|=1$, and $z \neq 0, z \neq 2 \mathrm{i}$. Also, let the complex number $\omega$ be such that $\frac{\omega-2 \mathrm{i}}{\omega} \cdot \frac{z}{z-2 \mathrm{i}}$ is a real number. Then the principal value of the argument of the complex number $\omega-2$ lies in the range $\qquad$.
Solve: Fill in $\left[\pi-\arctan \frac{4}{3}, \pi\right]$. Reason: Since $|z|=1$, the trajectory of point $z$ is a circle with $(0,1)$ as the center and 1 as the radius (excluding points $(0,0)$ and $(0,2)$), hence $\frac{z}{z-2 \mathrm{i}}$ is a pure imaginary number, and thus $\frac{\omega-2 \mathrm{i}}{\omega}$ is also a pure imaginary number. The trajectory of point $\omega$ is also a circle with $(0,1)$ as the center and 1 as the radius (excluding points $(0,0)$ and $(0,2)$). Therefore, $\arg (\omega-2) \in \left[\pi-\arctan -\frac{4}{3}, \pi\right]$.