8. In the complex plane, non-zero complex numbers $z_{1}, z_{2}$ lie on a circle centered at $\mathrm{i}$ 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 $z_{2}=$ $\qquad$ .
Solving
The equation of the circle with $\mathrm{i}$ as the center and 1 as the radius is $x^{2}+(y-1)^{2}=1$. It is easy to get $z_{1}=\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}$. Let $z_{2}=x+y \mathrm{i}$, then we have $\left\{\begin{array}{l}\operatorname{Re}\left[\left(\frac{\sqrt{3}}{2}-\frac{1}{2} \mathrm{i}\right)(x+y \mathrm{i})\right]=0, \\ x^{2}+(y-1)^{2}=1\end{array}\right.$
$$
\Rightarrow\left\{\begin{array}{l}
\frac{\sqrt{3}}{2} x+\frac{1}{2} y=0, \\
x^{2}+(y-1)^{2}=1
\end{array} \Rightarrow y^{2}+3(y-1)^{2}=3 \Rightarrow 4 y^{2}-6 y=0 \Rightarrow y=\frac{3}{2}\right.
$$
$\left(y=0\right.$ when $z_{2}=0$, discard $) \Rightarrow z_{2}=-\frac{\sqrt{3}}{2}+\frac{3}{2} \mathrm{i}$. Therefore, $z_{2}=-\frac{\sqrt{3}}{2}+\frac{3}{2} \mathrm{i}$.