For example, the vertices $B$ and $A$ of a regular $\triangle A B P$ with side length $a$ slide on the $x$-axis and $y$-axis respectively $(A, B, P$ are arranged clockwise). Find the trajectory of vertex $P$ in the first quadrant.
Let $P, A, B$ correspond to the complex numbers $x+yi, 2mi$, $2n$, respectively. Then the midpoint $C$ of $AB$ corresponds to the complex number $n+mi$, and $\overrightarrow{CP}=\overrightarrow{CB} \cdot \sqrt{3} \vec{i}$, i.e.,
$$
\begin{aligned}
& (x+yi)-(n+mi) \\
= & (2n-n-mi) \sqrt{3i} .
\end{aligned}
$$

Rearranging and applying the condition for equality of complex numbers, we get
$$
\left.n=\frac{1}{2}(\sqrt{3} y-x), m=\frac{1}{2} \sqrt{3 x}-y\right) \text {. }
$$

Given that $(2m)^{2}+(2n)^{2}=a^{2}$, we thus have
$$
\begin{array}{l}
x^{2}+y^{2}-\sqrt{3} xy=\frac{a^{2}}{4} . \quad\left(x \geqslant \frac{a}{2},\right. \\
\left.y \geqslant \frac{a}{2}\right)
\end{array}
$$
(Author's affiliation: Xiping High School, Henan)