One, (This section is 25 points) In the quadratic equation of $x$, $x^{2} + z_{1} x + z_{2} + m = 0$, $z_{1}, z_{2}, n$ are complex numbers, and $x_{1}^{2} - 4 z_{2} = 1 f + 20 i$.

Assume the two roots of this equation, $\alpha, \beta$, satisfy $|\alpha - \beta| = 2 \sqrt{7}$.
Find the maximum and minimum values of $|m|$.
$$
\begin{array}{l}
\left\{\begin{array}{l}
\alpha+\beta=-z_{1}, \\
\alpha \beta=z_{2}+m .
\end{array}\right. \\
\because(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m, \\
\therefore|\alpha-\beta|^{2}=\mid 4 m- \\
\left(z_{1}^{2}-4 z_{2}\right) \mid=28 . \\
\quad \therefore\left|m-\frac{1}{4}\left(z_{1}^{2}-4 z_{2}\right)\right| \\
=7,
\end{array}
$$
$$
\text { i.e., }|m-(4+5 i)|=7 \text {. }
$$

This indicates that the complex number $m$ lies on the circumference of a circle with center $A(4,5)$ and radius 7.
$\because|O A|=\sqrt{4^{2}+5^{2}}=\sqrt{41}<7$, hence the origin $O$ is inside $\odot A$. Connecting $O A$ and extending it to intersect $\odot A$ at points $B$ and $C$, then $|O B|=|O A|+|A B|=\sqrt{41}+7$ is the maximum value of $|m|$, and $|O C|=|C A|-|A O|=7-\sqrt{41}$ is the minimum value of $|m|$.
$\therefore|m|$ has a maximum value of $\sqrt{41}+7$, and a minimum value of 7- $\sqrt{41}$.
Solution two: As in solution one, we get $|m-(4+5 i)|=7$.
Let $m=x+y i$, then $\left\{\begin{array}{l}x=7 \cos \alpha+4, \\ y=7 \sin \alpha+5 .\end{array}\right.$
$$
\begin{array}{l}
|m|^{2}=x^{2}+y^{2}=90+56 \cos \alpha+70 \sin \alpha \\
=90+14 \sqrt{41}\left(\frac{1}{\sqrt{41}} \cos \alpha+\frac{5}{\sqrt{41}} \sin \alpha\right)
\end{array}
$$
$=90+14 \sqrt{41} \sin (\alpha+\varphi)$, where $\sin \varphi=\frac{4}{\sqrt{41}}$.
$\therefore|m|$ has a maximum value of $\sqrt{90+14 \sqrt{41}}=7+\sqrt{41}$,
$$
|m| \text { has a minimum value of }=\sqrt{90-14 \sqrt{41}}=7-\sqrt{41} \text {. }
$$