Example 2. Given the circle $x^{2}+y^{2}=4$ and a point $A(2, 0)$ on it, the moving chord $BC$ of this circle always satisfies the condition $\angle BAC = \frac{\pi}{3}$. Try to find the equation of the locus of the centroid of $\triangle ABC$.
Solve: Place the circle and $A$ on the corresponding complex plane, and let point $B$ correspond to the complex number $z^{\prime}$.
$$
\because \text{ the complex equation of the circle is } |z|=2
$$
$$
\therefore |z^{\prime}|=2 \text{, }
$$
$$
\begin{aligned}
& \because |B O| \\
= & |C O|,
\end{aligned}
$$
$$
\begin{array}{l}
\text { and } \angle B O C \\
=\frac{2 \pi}{8},
\end{array}
$$
$$
\therefore \text{ the complex number corresponding to point } C \text{ is }
$$
$$
\begin{array}{l}
z^{\prime}\left(\cos \frac{2 \pi}{8}+i \sin \frac{2 \pi}{8}\right) \\
=\left(-\frac{1}{2}+\frac{\sqrt{8}}{2} i\right) z^{\prime} .
\end{array}
$$

Let the centroid $G$ of $\triangle A B C$ correspond to the complex number $\mathrm{z}$,
$$
\begin{array}{l} 
\text { then } \mathrm{z}=\frac{z_{\mathrm{A}}+z_{\mathrm{B}}+z_{\mathrm{C}}}{3} \\
= \frac{1}{3}\left[2+z^{\prime}+\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) z^{\prime}\right] \\
\Rightarrow z^{\prime}\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=3 z-2 .
\end{array}
$$

Taking the modulus on both sides of (2) and substituting (1) we get
$$
|3 z-2|=2 \Rightarrow\left|z-\frac{2}{3}\right|=\frac{2}{3} \text {. }
$$

Converting to Cartesian coordinates and considering the purity of the trajectory, the arc of the circle $\left(x-\frac{2}{3}\right)^{2}+y^{2}=\left(\frac{3}{2}\right)^{2}$ to the left of the line $x=1$ is the desired result.