Example 8 In the quadratic equation $x^{2}+z_{1} x+z_{2}+m=0$ about $x$, $z_{1}, z_{2}, m$ are all complex numbers, and $z_{1}^{2}-$ $4 z_{2}=16+20 \mathrm{i}$. Let the two roots of this equation be $\alpha, \beta$, satisfying $|\alpha-\beta|=2 \sqrt{7}$, find the maximum and minimum values of $|m|$.
From Vieta's formulas, we get $\alpha+\beta=-z_{1}, \alpha \beta=z_{2}+m$.
Since $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m$.
Therefore, $|\alpha-\beta|^{2}=|4 m-(z_{1}^{2}-4 z_{2})|=28$.
That is, $|m-(4+5 i)|=7$.
This indicates that the complex number $m$ lies on the circumference of a circle with center $A(4,5)$ and radius 7, and the origin $O$ is inside the circle $\odot A$. As shown in Figure 20-1, connect $O A$ and extend it to intersect $\odot A$ at points $B$ and $C$. It is easy to solve that
$$
\begin{array}{l}
|m|_{\text {max }}=|O B|=|O A|+|A B|=\sqrt{41}+7 . \\
|m|_{\text {min }}=|O C|=|C A|-|A O|=7-\sqrt{41} .
\end{array}
$$