Example 10 (1996 National High School League Question) On the complex plane, non-zero complex numbers $z_{1}$ and $z_{2}$ lie on a circle with center at $\mathrm{i}$ (the corresponding point) and radius 1. The real part of $\bar{z}_{1} \cdot z$ is zero, and the principal value of the argument of $z_{1}$ is $\frac{\pi}{6}$, then $z_{2}=$ $\qquad$ .
Solve: Fill in $-\frac{\sqrt{3}}{2}+\frac{3}{2} \mathrm{i}$. Reason: Since $\arg z_{1}=\frac{\pi}{6}$, the vector $\overrightarrow{O Z}$ corresponding to $z_{1}$ forms an angle of $\frac{\pi}{3}$ with the $y$-axis, so $\left|z_{1}\right|=1, z_{1}=\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}$.

Also, the real part of $\bar{z}_{1} \cdot z_{2}$ is zero, so $\arg z_{2}-\frac{\pi}{6}=\frac{\pi}{2}$ or $\frac{3}{2} \pi$. Since $z_{2}$ lies on the circle $|z-\mathrm{i}|=1$, $\frac{3}{2} \pi$ should be discarded, i.e., $\arg z_{2}=\frac{2}{3} \pi$. $z_{2}$ lies on the diameter passing through $z_{1}$ and the center of the circle, $\left|z_{2}\right|=\sqrt{3}$,
thus $z_{2}=\sqrt{3}\left(\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}\right)=-\frac{\sqrt{3}}{2}+\frac{3}{2} \mathrm{i}$.