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 $\overline{z_{1}} \cdot z_{2}$ is zero, and the principal value of the argument of $z_{1}$ is $\frac{\pi}{6}$. Then $z_{2}=$ $\qquad$ .
2. $z_{2}=-\frac{\sqrt{3}}{2}+\frac{3}{2} i$.

Since $\arg z_{1}=\frac{\pi}{6}, z_{1}$ forms an angle of $\frac{\pi}{3}$ with the $y$-axis,
thus, $\left|z_{1}\right|=1$,
$$
z_{1}=\frac{\sqrt{3}}{2}+\frac{1}{2} i \text {. }
$$

Also, the real part of $\overline{z_{1}} \cdot z_{2}$ is zero,
$$
\therefore \arg z_{2}-\frac{\pi}{6}=\frac{\pi}{2} \text { or } \frac{3}{2} \pi \text {. }
$$

Since $z_{2}$ lies on the circle $|z-i|=1$, $\frac{3}{2} \pi$ should be discarded, so $\arg z_{2} = \frac{2}{3} \pi$.
$z_{2}$ lies on the diameter passing through $z_{1}$ and the center of the circle, $\left|z_{2}\right|=\sqrt{3}$, hence $z_{2}=\sqrt{3}\left(\cos \frac{2}{3} \pi+i \sin \frac{2 \pi}{3}\right)$
$$
=-\frac{\sqrt{3}}{2}+\frac{3}{2} i \text {. }
$$