Example 1 Given that $a, b, c, d$ take certain real values, the equation $x^{4} + a x^{3} + b x^{2} + c x + d = 0$ has 4 non-real roots, where the product of 2 of the roots is $13 + \mathrm{i}$, and the sum of the other 2 roots is $3 + 4 \mathrm{i}$, where $\mathrm{i}$ is the imaginary unit. Find $b$.
(13th American Invitational Mathematics Examination)
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the 4 roots of the equation. By the problem, we can assume $x_{3}=\overline{x_{1}}, x_{4}=\overline{x_{2}}$, then we have
$$
\begin{aligned}
x_{1} x_{2} & =13+\mathrm{i}, x_{3}+x_{4}=3+4 \mathrm{i}. \text{ Hence } \\
b & =x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4} \\
& =\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+x_{1} x_{2}+x_{3} x_{4} \\
& =(3-4 \mathrm{i})(3+4 \mathrm{i})+(13+\mathrm{i})+(13-\mathrm{i}) \\
& =51 .
\end{aligned}
$$