Example 4. When $\mathrm{m}$ is what real number, the argument $\theta=\pi+\frac{\pi}{4}$ of the complex number $\left(\mathrm{m}^{2}-\mathrm{m}\right.$ $-2)+\left(2 \mathrm{~m}^{2}-3 \mathrm{~m}-2\right) \mathrm{i}$ ? Find the modulus of this complex number.
Solve: From the conditions, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
m^{2}-m-2<0, \\
2 m^{2}-3 m-2<0, \\
m^{2}-m-2=2 m^{2}-3 m-2 .
\end{array}\right. \\
\text { if }\left\{\begin{array}{l}
-1<m<2, \\
-\frac{1}{2}<m<2, \\
m=0, m=2 .
\end{array}\right. \\
\end{array}
$$

Therefore, when $m=0$, the argument of the complex number $\theta=\pi+\frac{\pi}{4}$.
This complex number is $-2-2i$, and its modulus $|-2-2i|=$ $=\sqrt{(-2)^{2}+(-2)^{2}}=2\sqrt{2}$.