4. Let $|a|=1,|b|=2$. If vector $c$ satisfies
$$
|c-(a+b)|=|a-b|,
$$

then the maximum value of $|\boldsymbol{c}|$ is $\qquad$
4. $2 \sqrt{5}$.
$$
\begin{array}{l}
\text { Given }|\boldsymbol{c}-(\boldsymbol{a}+\boldsymbol{b})|^{2}=|a-b|^{2} \\
\Rightarrow(\boldsymbol{c}-2 a)(c-2 b)=0 .
\end{array}
$$

Let the starting point of the vectors be the origin $O, \overrightarrow{O A}=2 a$, $\overrightarrow{O B}=2 \boldsymbol{b}, \overrightarrow{O C}=c$. Then points $A$ and $B$ lie on the circumference of circles centered at $O$ with radii 2 and 4, respectively.
By symmetry, assume point $C$ lies on the $x$-axis.
To satisfy equation (1), i.e., $\angle A C B=90^{\circ}$. To maximize $|\boldsymbol{c}|=$ $|\overrightarrow{O C}|=x$, $A C$ and $B C$ must be tangents to the two circles, as shown in Figure 3.
Notice that,
$$
\begin{array}{l}
\cos \angle A O B=\cos (\angle A O C+\angle B O C) \\
=\cos 90^{\circ}=0 .
\end{array}
$$

Solving this, we get $x=2 \sqrt{5}$.