Example 1. For a regular $\triangle ABC$, the vertex $B$ has coordinates $(m, 0)$ $\left(m > a \text{ is a constant}\right)$, and point $A$ moves along the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$. Assuming $A$, $B$, and $C$ are arranged counterclockwise, find the equation of the trajectory of point $C$.
$$
\begin{array}{l}
\text { Let } A, B, C \text { be the points corresponding to the complex numbers } x_{0}+y_{0} i \text {, } \\
\rightarrow \\
m, x+y i \text {, then } B A_{1}\left(x_{0}-m\right)+y_{0} i \text {, } \\
B C: \quad(x-m)+y i \text {. } \\
\end{array}
$$

Since $\overrightarrow{B C}$ rotated counterclockwise by $\frac{\pi}{3}$ gives $\overrightarrow{B A}$, i.e., $\overrightarrow{B A}=\overrightarrow{B C} e^{-\vec{s}:}$,
using the definition of equality of complex numbers, we get:
$$
\begin{array}{l}
x_{0}-m=\frac{1}{2}(x-m)-\frac{\sqrt{3}}{2} y, \\
y_{0}=\frac{y}{2}+\frac{\sqrt{3}}{2}(x-m) .
\end{array}
$$

But $A\left(x_{0}, y_{0}\right)$ lies on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,
solving for $x_{0}, y_{0}$ and substituting, we get the trajectory equation of point $C$:
$$
\begin{array}{l}
\frac{(x+m-\sqrt{3} y)^{2}}{4 a^{2}} \\
+\frac{[y+\sqrt{3}(x-m)]^{2}}{4 b^{2}}=1
\end{array}
$$