5. In the Cartesian coordinate system $x O y$, the moving point $A$ is on the circle $x^{2}+y^{2}=1$, and the coordinates of point $B$ are $(3,0)$. If point $C$ makes $\triangle A B C$ an equilateral triangle, then the maximum value of $|O C|$ is $\qquad$.
Given:
$$
z_{A}=\cos \theta+\mathrm{i} \sin \theta, \quad z_{B}=3,
$$
then
$$
\begin{array}{l}
z_{\overrightarrow{B A}}=(\cos \theta-3)+\mathrm{i} \sin \theta \\
\Rightarrow z_{\overrightarrow{B C}}=[(\cos \theta-3)+\mathrm{i} \sin \theta] \cdot\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right) \\
=\cos \left(\theta-\frac{\pi}{3}\right)-\frac{3}{2}+\left[\sin \left(\theta-\frac{\pi}{3}\right)+\frac{3 \sqrt{3}}{2}\right] \mathrm{i} .
\end{array}
$$

Thus, the trajectory of point $C$ is a circle with center $M\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)$ and radius 1, so the maximum value of $|O C|$ is $3+1=4$.