9. Given the complex sequence $\left\{z_{n}\right\}$ satisfies: $z_{1}=\frac{\sqrt{3}}{2}, z_{n+1}=\overline{z_{n}}\left(1+z_{n} \mathrm{i}\right)(n=1,2, \cdots)$, where $\mathrm{i}$ is the imaginary unit, find the value of $z_{2021}$.
Let $z_{n}=a_{n}+b_{n} \mathrm{i}$, where $a_{n}, b_{n} \in \mathbf{R}$, then $a_{n+1}+b_{n+1} \mathrm{i}=a_{n}-b_{n} \mathrm{i}+\left(a_{n}^{2}+b_{n}^{2}\right) \mathrm{i}$, thus $a_{n+1}=a_{n}, b_{n+1}=a_{n}^{2}+b_{n}^{2}-b_{n}$. Combining $a_{1}=\frac{\sqrt{3}}{2}, b_{1}=0$ we get
$$
\begin{array}{l}
a_{n}=\frac{\sqrt{3}}{2}, b_{n+1}=b_{n}^{2}-b_{n}+\frac{3}{4} \Rightarrow b_{n+1}-\frac{1}{2}=\left(b_{n}-\frac{1}{2}\right)^{2} \\
\Rightarrow b_{n}-\frac{1}{2}=\left(b_{1}-\frac{1}{2}\right)^{2^{n-1}} \Rightarrow b_{n}=\frac{1}{2}+\frac{1}{2^{2 n-1}}\left(n \in \mathbf{N}^{*}\right) . \\
\text { In summary, } z_{2021}=\frac{\sqrt{3}}{2}+\left(\frac{1}{2}+\frac{1}{2^{20200}}\right) \mathrm{i} .
\end{array}
$$