4. Let $a, b$ be complex numbers, their conjugate complex numbers are $\bar{a}, \bar{b}$, and their moduli are $|a|, |b|$. If $a+2 \bar{b}=\mathrm{i}$, $\bar{a} \cdot \bar{b}=-5-\mathrm{i}$, then $|a|^{2}$ is $\qquad$ .
4.8 or 13.

From $2 \bar{a} \cdot \bar{b}=-10-2 \mathrm{i}$ and $a+2 \bar{b}=\mathrm{i}$, eliminating $2 \bar{b}$, we get $\bar{a}(\mathrm{i}-a)=-10-2 \mathrm{i}$,
Taking the conjugate of both sides of the above equation, we get $a(-\mathrm{i}-\bar{a})=-10+2 \mathrm{i}$, which is
$$
\begin{array}{l}
\text { i } \bar{a}-\bar{a} \cdot a=-10-2 \mathrm{i}, \\
-\mathrm{i} a-\bar{a} \cdot a=-10+2 \mathrm{i} .
\end{array}
$$
(1) - (2) gives
$$
\mathrm{i}(\bar{a}+a)=-4 \mathrm{i}, \bar{a}+a=-4 \text {, }
$$

which means $\operatorname{Re}(a)=-2$.
Also, (1) + (2) and rearranging gives
$$
\mathrm{i}(a-\bar{a})=20-2 \bar{a} \cdot a \text {, }
$$

which means $\operatorname{Im}(a)=\frac{20-2|a|^{2}}{-2}=|a|^{2}-10$.
Thus, $|a|^{2}=(\operatorname{Re}(a))^{2}+(\operatorname{Im}(a))^{2}$
$$
=|a|^{4}-20|a|^{2}+104 \text {, }
$$

which simplifies to $\left(|a|^{2}\right)^{2}-21|a|^{2}+104=0$.
Solving this, we get $|a|^{2}=8$ or 13.