Example 1. Given $A(-a, 0), B(a, 0)$, a moving point $M$ is on the circle $x^{2}+y^{2}=r^{2}$, $a>r>0$, and equilateral triangles $\mathrm{AMC}$ and BMD are constructed above the broken line AMB with $AM$ and $BM$ as sides, respectively. Find the equation of the trajectory of the midpoint $\mathrm{P}$ of $C D$.
Consider the x-axis as the real axis in the complex plane, and the y-axis as the imaginary axis. Let $Z_{\boldsymbol{\mu}}=r(\cos \theta + i \sin \theta) \cdot (0 \leqslant \theta < 2 \pi)$

$\overrightarrow{M_{1}}$ corresponds to the complex number: $(r \cos \theta - a) + ir \sin \theta$,

$\vec{A} M$ corresponds to the complex number: $(r \cos \theta + a) + ir \sin \theta$.

\[
\begin{aligned}
\overrightarrow{A C} & = \overrightarrow{A M} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \\
& = \left[ (r \cos \theta + a) + i r \sin \theta \right] \left( \frac{1}{2} + \frac{\sqrt{3}}{2} i \right) \\
& = \frac{r \cos \theta + a - \sqrt{3} r \sin \theta}{2} \\
& + \frac{r \sin \theta + \sqrt{3} r \cos \theta + \sqrt{3} a}{2} i .
\end{aligned}
\]

$\overrightarrow{\text{Since}} = Z_{0} - Z_{A}$, hence $Z_{C} = \frac{r \cos \theta - a - \sqrt{3} r \sin \theta}{2}$

\[
\begin{array}{l}
+ \frac{r \sin \theta - \sqrt{3} r \cos \theta + \sqrt{3} a}{2} i.
\end{array}
\]

Let $Z_{k} = x + y i (x, y \in R)$, then

\[
\begin{array}{l}
x = \frac{r \cos \theta}{2}, \\
y = \frac{r \sin \theta}{2} + \frac{\sqrt{3} a}{2}.
\end{array}
\]

Eliminating the parameter, we get $x^{2} + \left( y - \frac{\sqrt{3}}{2} a \right)^{2} = \frac{r^{2}}{4}$.

The required trajectory is a circle with center $\left( 0, \frac{\sqrt{3} a}{2} \right)$ and radius $\frac{r}{2}$.