4. Let $x, y$ be complex numbers. Then
$$
\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}}
$$

the maximum value is . $\qquad$
4. $\frac{5 \sqrt{2}}{2}$.
First, prove: $\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}} \leqslant \frac{5 \sqrt{2}}{2}$.
It suffices to prove:
$$
|3 x+4 y|^{2} \leqslant \frac{25}{2}\left(|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|\right) \text {, }
$$

which is
$$
\begin{array}{l}
2(3 x+4 y)(3 \bar{x}+4 \bar{y})-25\left(|x|^{2}+|y|^{2}\right) \\
\leqslant 25\left|x^{2}+y^{2}\right|,
\end{array}
$$

which is $\square$
$$
\begin{array}{l}
-7|x|^{2}+7|y|^{2}+24(\bar{x} y+x \bar{y}) \\
\leqslant 25\left|x^{2}+y^{2}\right| .
\end{array}
$$

It suffices to prove:
$$
\begin{array}{l}
{[7(-x \bar{x}+y \bar{y})+24(\bar{x} y+x y)]^{2}} \\
\leqslant 625\left(x^{2} \dot{x}^{2}+y^{2} \bar{y}^{2}+\bar{x}^{2} y^{2}+x^{2} \bar{y}^{2}\right) .
\end{array}
$$

Noting that $\bar{x} y+x \bar{y}=\overline{\bar{x} y+x \bar{y}}$ is a real number, then by the Cauchy-Schwarz inequality we have
$$
\begin{array}{l}
{[7(-x \bar{x}+y \bar{y})+24(1+x \bar{y})]^{2}} \\
\leqslant\left(7^{2}+24^{2}\right)\left[(-x \bar{x}+\bar{y})^{2}+(\bar{x} y+x \bar{y})^{2}\right] \\
=625\left(x^{2} \bar{x}^{2}+y^{2} \bar{y}^{2}+\bar{x}^{2} y^{2}+x^{2} \bar{y}^{2}\right) .
\end{array}
$$
Equality in (1) holds when $x=1, y=\frac{4}{3}$.