4. If the complex coefficient equation $(4+3 \mathrm{i}) x^{2}+m x+$ $4-3 \mathrm{i}=0$ has real roots, then the minimum value of the modulus of the complex number $m$ is $\qquad$
4. 8 .

Let $\alpha$ be a real root of the original equation, $m=x+y \mathrm{i}$, then
$$
(4+3 \mathrm{i}) \alpha^{2}+(x+y \mathrm{i}) \alpha+4-3 \mathrm{i}=0,
$$

which means
$$
\left\{\begin{array}{l}
4 \alpha^{2}+\alpha x+4=0, \\
3 \alpha^{2}+\alpha y-3=0,
\end{array}\right.
$$

or equivalently
$$
\left\{\begin{array}{l}
x=-4\left(\alpha+\frac{1}{\alpha}\right), \\
y=-3\left(\alpha-\frac{1}{\alpha}\right) .
\end{array}\right.
$$

Thus,
$$
|m|^{2}=x^{2}+y^{2}=16\left(\alpha+\frac{1}{\alpha}\right)^{2}+9\left(\alpha-\frac{1}{\alpha}\right)^{2}=25\left(\alpha^{2}+\frac{1}{\alpha^{2}}\right)+14 \geqslant 64,
$$

which implies $|m| \geqslant 8$. The equality holds if and only if $\alpha^{4}=1$, i.e., $\alpha= \pm 1$, in which case $m= \pm 8$. Therefore, the minimum value of the modulus of $m$ is 8.