11. First, prove
$$
\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}} \leqslant \frac{5 \sqrt{2}}{2} \text {. }
$$

It suffices to prove
$$
|3 x+4 y|^{2} \leqslant \frac{25}{2}\left(|x|^{2}+|y|^{2}+|x+y|^{2}\right),
$$

which is equivalent to
$$
2(3 x+4 y)(3 \bar{x}+4 \bar{y})-25\left(|x|^{2}+|y|^{2}\right) \leqslant 25\left|x^{2}+y^{2}\right|,
$$

which simplifies to
$$
-7|x|^{2}+7|y|^{2}+24(\bar{x} y+x \bar{y}) \leqslant 25\left|x^{2}+y^{2}\right| .
$$

It suffices to prove
$$
(7(-x \bar{x}+y \bar{y})+24(\bar{x} y+x \bar{y}))^{2} \leqslant 625\left(x^{2} \bar{x}^{2}+y^{2} \bar{y}^{2}+\bar{x}^{2} y^{2}+x^{2} \bar{y}^{2}\right) .
$$

Notice that $\bar{x} y+x \bar{y}=\overline{\bar{x}} y+x \bar{y}$ is a real number. By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
(7(-x \bar{x}+y \bar{y})+24(\bar{x} y+x \bar{y}))^{2} & \leqslant\left(7^{2}+24^{2}\right)\left((-x \bar{x}+y \bar{y})^{2}+(\bar{x} y+x \bar{y})^{2}\right) \\
& =625\left(x^{2} \bar{x}^{2}+y^{2} \bar{y}^{2}+\bar{x}^{2} y^{2}+x^{2} \bar{y}^{2}\right) .
\end{aligned}
$$

The equality holds when $x=1, y=\frac{4}{3}$. Therefore, the maximum value of $\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}}$ is $\frac{5 \sqrt{2}}{2}$.
11. (20 points) Let $x, y$ be complex numbers, find the maximum value of $\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}}$.

