Question 164, Point $\mathrm{P}$ moves on the circle $(\mathrm{x}-2)^{2}+(\mathrm{y}-1)^{2}=1$, vector $\overrightarrow{\mathrm{PO}}$ (where $\mathrm{O}$ is the origin of coordinates) rotates counterclockwise by $90^{\circ}$ around point $\mathrm{P}$ to get $\overrightarrow{\mathrm{PQ}}$, then the trajectory equation of point $\mathrm{Q}$ is $\qquad$ _.
Question 164, Analysis: Let point $\mathrm{P}(2+\cos \theta, 1+\sin \theta)$. If we consider this plane as the complex plane, then the complex number corresponding to point $\mathrm{P}$ is $(2+\cos \alpha)+(1+\sin \alpha) \mathrm{i}$. Therefore, the complex number corresponding to point $Q$ is
$$
\begin{array}{l}
{[-(2+\cos \alpha)-(1+\sin \alpha) i] i+[(2+\cos \alpha)+(1+\sin \alpha) i]} \\
=(3+\cos \alpha+\sin \alpha)+(-1-\cos \alpha+\sin \alpha) i
\end{array}
$$

Thus, we know that $Q(3+\cos \alpha+\sin \alpha,-1-\cos \alpha+\sin \alpha)$, which means
$$
(x-3)^{2}+(y+1)^{2}=(\cos \alpha+\sin \alpha)^{2}+(-\cos \alpha+\sin \alpha)^{2}=2
$$

Therefore, the trajectory of point $Q$ is $(x-3)^{2}+(y+1)^{2}=2$.