Example 2 Let the vertices of $\triangle ABC$ correspond to the complex numbers $z_{1}, z_{2}, z_{3}$, and satisfy $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1$. Try to express the orthocenter and the center of the nine-point circle of $\triangle ABC$ using $z_{1}, z_{2}, z_{3}$.
Analysis: Let $O, H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively. Then the nine-point circle (the midpoints of the three sides, the three vertices of the pedal triangle, and the midpoints of the segments connecting the orthocenter to each vertex, these nine points are concyclic) has its center at the midpoint of $OH$. Given that $O$ is the origin of the complex plane, we only need to find the complex number corresponding to $H$. We can use the necessary and sufficient condition for perpendicularity to calculate.

Solution: Let $O, H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively, and let the complex number corresponding to $H$ be $z$. Then $\frac{z-z_{1}}{z_{3}-z_{2}}, \frac{z-z_{2}}{z_{3}-z_{1}}$ are both purely imaginary numbers. Therefore, $\frac{z-z_{1}}{z_{3}-z_{2}}=-\frac{\bar{z}-\bar{z}_{1}}{\bar{z}_{3}-\bar{z}_{2}}, \frac{z-z_{2}}{z_{3}-z_{1}}=-\frac{\bar{z}-\bar{z}_{2}}{\bar{z}_{3}-\bar{z}_{1}}$. Eliminating $\bar{z}$ and combining with $z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=z_{3} \bar{z}_{3}=1$, we get $z=z_{1}+z_{2}+z_{3}$.

Noting that the center of the nine-point circle of $\triangle ABC$ is the midpoint of $OH$, and $O$ is the origin of the complex plane, the complex number corresponding to the center of the nine-point circle is $\frac{1}{2}\left(z_{1}+z_{2}+z_{3}\right)$.