8. In the complex plane, points $A, B$ correspond to the complex numbers $z_{1}=2, z_{2}=-3$, respectively, and point $P$ corresponds to the complex number $z$. The principal value of the argument of $\frac{z-z_{1}}{z-z_{2}}$ is $\varphi$. When point $P$ moves on the upper semicircle (excluding the two endpoints) with the origin as the center and 1 as the radius, find the minimum value of $\varphi$.
8. As shown in Figure $18-5$. Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, then $x^{2}+y^{2}=1(y>0)$.
Let $\arg \left(z-z_{1}\right)=\alpha, \arg \left(z-z_{2}\right)=\beta$, then
$$
\varphi=\alpha-\beta \text {, and } \frac{\pi}{2}<\varphi<\pi \text {. }
$$

Since $\tan \alpha=\frac{y}{x-2}, \tan \beta=\frac{y}{x+3}$,
we have $\tan \varphi=\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}=\frac{5 y}{x-5}$.
According to (1), we can set $x=\cos \theta, y=\sin \theta, \theta \in(0, \pi)$, then $\tan \varphi=\frac{5 \sin \theta}{\cos \theta-5}$. That is, $5 \sin \theta-\tan \varphi \cdot \cos \theta=-5 \tan \varphi$.
The necessary and sufficient condition for equation (3) to have real solutions in $\theta \in(0, \pi)$ is $\left|\frac{-5 \tan \varphi}{\sqrt{5^{2}+\tan ^{2} \varphi}}\right| \leqslant 1$.
Solving this, we get $-\frac{5 \sqrt{6}}{12} \leqslant \tan \varphi \leqslant \frac{5 \sqrt{6}}{12}$.
From (2) and (4), we get $\varphi_{\text {min }}=\pi-\arctan \frac{5 \sqrt{6}}{12}$.
At this point, the coordinates of $P$ are $\left(\frac{1}{5}, \frac{2 \sqrt{6}}{5}\right)$.