Example 3 In the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, there is a point $M(P, 0)$, and $0<P<a$, and a moving point $Q$ on the ellipse. The vector $\overrightarrow{M Q}$ rotates $90^{\circ}$ clockwise around point $M$ to reach $\overrightarrow{M P}$. Try to find the equation of the trajectory of point $P$.
Analyzing the use of complex numbers to implement vector rotation transformations is a good choice.
Solution: Let the coordinates of the moving point $Q$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ be $(a \cos \theta, b \sin \theta)$, then the complex number corresponding to $\overrightarrow{M Q}$ is $(a \cos \theta-P)+(b \sin \theta) \mathrm{i}$.
$\overrightarrow{M Q}$ is rotated $90^{\circ}$ clockwise around point $M$ to $\overrightarrow{M P}$, and let the coordinates of point $P$ be $(x, y)$, thus
$$
(a \cos \theta-P+b \sin \theta)(-\mathrm{i})=x-P+y \mathrm{i} \text {. }
$$

That is, $b \sin \theta+(P-a \cos \theta) \mathrm{i}=x-P+y \mathrm{i}$. Therefore,
$$
\left\{\begin{array}{l}
x-P-b \sin \theta \\
y=P-a \cos \theta
\end{array}\right.
$$

Eliminating the parameter $\theta$, we obtain the trajectory equation of point $P$ as $\frac{(x-P)^{2}}{b^{2}}+\frac{(y-P)^{2}}{a^{2}}=1$.