Think about question 2: In the complex plane, points $A$ and $B$ correspond to the complex numbers $z_{1}=2$, $z_{2}=-3$, respectively. Point $P$ corresponds to the complex number $z$, and the principal value of the argument of $\frac{z-z_{1}}{z-z_{2}}$ is $\varphi$. When point $P$ moves on the upper semicircle with the origin as the center and a radius of 1 (excluding the two endpoints), find the minimum value of $\varphi$.
Let $z=\cos \theta+i \sin \theta, \theta \in(0, \pi)$. As shown in Figure 2-7, let $\arg \left(z-z_{1}\right)=\alpha, \arg \left(z-z_{2}\right)=\beta$. Then, by the geometric meaning of complex number division, we have $\varphi=\alpha-\beta$, and $\varphi \in\left(\frac{\pi}{2}, \pi\right)$.
Given $\tan \alpha=\frac{\sin \theta}{\cos \theta-2}, \tan \beta=\frac{\sin \theta}{\cos \theta+3}$, we get
$$
\tan \varphi=\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{5 \sin \theta}{\cos \theta-5},
$$

Therefore, we have $5 \sin \theta-\tan \varphi \cdot \cos \theta=-5 \tan \varphi$.
Thus, $\sqrt{5^{2}+(-\tan \varphi)^{2}} \geqslant|-5 \tan \varphi|$,
Solving this, we get $|\tan \varphi| \leqslant \frac{5 \sqrt{6}}{12}$. Since $\frac{\pi}{2}<\varphi<\pi$, we have $\varphi \geqslant \pi-\arctan \frac{5 \sqrt{6}}{12}$,
The equality holds when $z=\frac{1}{5}+\frac{2 \sqrt{6}}{5} i$. Therefore, the minimum value of $\varphi$ is $\pi-\arctan \frac{5 \sqrt{6}}{12}$.