2. In the complex plane, non-zero complex numbers $z_{1}, z_{2}$ lie on a circle centered at $i$ with a radius of 1, the real part of $z_{1} z_{2}$ is zero, and the principal value of the argument of $z_{1}$ is $\frac{1}{6} \pi$, then $z_{2}=$ $\qquad$
$$
-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} i
$$
2. 【Analysis and Solution】 The real part of $\overline{z_{1}} z_{2}$ is 0, so $\arg \left(\overline{z_{1}} z_{2}\right)=\frac{\pi}{2}$, or $\frac{3}{2} \pi$, which means $\arg z_{2}-\frac{\pi}{6}=\frac{\pi}{2}$ or $\arg z_{2}-$ $\frac{\pi}{6}=\frac{3}{2} \pi$. Since $z_{2}$ is on the specified circle, the argument cannot be greater than $\pi$. Therefore, $\arg z_{2}=$ $\frac{2}{3} \pi$.

As shown in the figure, $\angle O_{1} O Z_{2}=30^{\circ}$, it is easy to know that the inscribed angle subtended by $O Z_{2}$ is $60^{\circ}$, so $O Z_{2}=\sqrt{3}$, i.e., $\left|Z_{2}\right|=\sqrt{3}$,
Therefore, $Z_{2}=\sqrt{3}\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)=-\frac{\sqrt{3}}{2}+\frac{3}{2} i$.