Example 3. Let points $A$ and $B$ slide on the $x$-axis and $y$-axis respectively, and $|A B|=a$. Construct an isosceles triangle with $A B$ as the base. Find the equation of the locus of point $C$.

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Rhodium $A, B, C$ correspond to the complex numbers $x^{\prime}$, $y^{\prime} i, x+i y\left(x^{\prime}, y^{\prime}, x, y \in R\right)$, respectively. Therefore, $\overrightarrow{B A}$ and $\overrightarrow{B C}$ correspond to the complex numbers $x^{\prime}-y^{\prime} i, x+\left(y-y^{\prime}\right) i$
$$
\begin{array}{l}
\because|A B|=a, \\
\therefore\left|x^{\prime}-y^{\prime}\right|=0
\end{array}
$$

That is, $x^{\prime 2}+y^{\prime 2}=a^{2}$.
Furthermore, $\overrightarrow{B C}$ can be obtained by rotating $\overrightarrow{B A}$ counterclockwise around $B$ without changing its length, hence
$$
x+\left(y-y^{\prime}\right) i=\left(x-y^{\prime} i\right) i=y^{\prime}+x^{\prime} i_{0}
$$

By the condition for equality of two complex numbers, we get $\left\{\begin{array}{l}y^{\prime}=x, \\ y-y^{\prime}=x^{\prime} .\end{array}\right.$
Thus, $\left\{\begin{array}{l}x^{\prime}=y-x \\ y^{3}=x .\end{array}\right.$
Squaring and adding these equations and substituting into (1) yields
$$
(x-y)^{2}+y^{2}=a^{2} \text {, }
$$