4.206 There are two forces $f_{1}$ and $f_{2}$ acting on the origin $O$ of the coordinate axis,
$$\begin{array}{l}
\vec{f}_{1}=\overrightarrow{O A}=\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
\vec{f}_{2}=\overrightarrow{O B}=2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right]
\end{array}$$
(1) Find the magnitude and direction of their resultant force;
(2) Find the distance between points $A$ and $B$ (accurate to 0.1).
[Solution]
$$\text { (1) Resultant force } \begin{aligned}
\vec{f}= & \overrightarrow{f_{1}}+\overrightarrow{f_{2}} \\
= & \sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
& +2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right] \\
= & 1+i+\sqrt{3}-i \\
= & \sqrt{3}+1 \\
= & (\sqrt{3}+1)\left(\cos 0^{\circ}+i \sin 0^{\circ}\right) .
\end{aligned}$$

So the magnitude of the resultant force is $\sqrt{3}+1$, and its direction is the same as the $x$-axis.
(2) Since $\overrightarrow{O A}=1+i$, the Cartesian coordinates of point $A$ are $(1,1)$, and $\overrightarrow{O B}=\sqrt{3}-i$, so the Cartesian coordinates of point $B$ are $(\sqrt{3},-1)$.
$$\text { Therefore } \begin{aligned}
|A B| & =\sqrt{(\sqrt{3}-1)^{2}+(-1-1)^{2}} \\
& =\sqrt{8-2 \sqrt{3}} \approx 2.1
\end{aligned}$$