Example 4 In the complex plane, point $A$ corresponds to the complex number -2, and moving point $B$ is on a circle centered at the origin with a radius of 1. A regular triangle $ABC$ (with $A, B, C$ in counterclockwise order) is constructed with $AB$ as one side. Find the trajectory of moving point $C$.
Analyzing the condition for triangle $ABC$ to be an equilateral triangle $(A, B, C$ arranged counterclockwise $)$ is that rotating vector $\overrightarrow{AB}$ counterclockwise by $\frac{\pi}{3}$ around point $A$ results in vector $\overrightarrow{AC}$, i.e., $\overrightarrow{AB} \cdot\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=\overrightarrow{AC}$. From this, it is not difficult to derive the trajectory equation of the moving point $C$.

Solution As shown in Figure 2-2, let the complex number corresponding to the moving point $C$ be $z$, and the complex numbers corresponding to points $A$ and $B$ be $z_{A}$ and $z_{B}$, respectively, with $z_{A}=-2$ and $\left|z_{B}\right|=1$.

By the given condition, $\triangle ABC$ is an equilateral triangle, so rotating vector $\overrightarrow{AB}$ counterclockwise by $\frac{\pi}{3}$ around point $A$ results in vector $\overrightarrow{AC}$, i.e., $\overrightarrow{AB} \cdot\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)=\overrightarrow{AC}$,

Thus, $\left(z_{B}-z_{A}\right)\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)=z-z_{A}$.
So, $z_{B}=\left(z-z_{A}\right)\left[\cos \left(-\frac{\pi}{3}\right)+\mathrm{i} \sin \left(-\frac{\pi}{3}\right)\right]+z_{A}$. Taking the modulus on both sides, and given $\left|z_{B}\right|=1$ and $z_{A}=-2$,
we get $\left|\left[\cos \left(-\frac{\pi}{3}\right)+\mathrm{i} \sin \left(-\frac{\pi}{3}\right)\right] \cdot\left[(z+2)-2\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)\right]\right|=1$.
This simplifies to $|z-(-1+\sqrt{3} \mathrm{i})|=1$. Therefore, the trajectory of the moving point $C$ is a circle with center at $(-1, \sqrt{3})$ and radius 1.