Question 136, Given the complex sequence $\left\{z_{n}\right\}$ satisfies: $z_{1}=\frac{\sqrt{3}}{2}, z_{n+1}=\overline{z_{n}}\left(1+z_{n} i\right)(n=1, 2, \ldots)$, where $\mathrm{i}$ is the imaginary unit. Find the value of $\mathrm{z}_{2021}$.
Question 136, Solution: Let $z_{n}=a_{n}+b_{n} i$, where $a_{n}, b_{n} \in \mathbb{R}$. Then $a_{1}=\frac{\sqrt{3}}{2}, b_{1}=0$, and according to the conditions,
$$
\begin{array}{l}
a_{n+1}+b_{n+1} i=\left(a_{n}-b_{n} i\right)\left(1+\left(a_{n}+b_{n} i\right) i\right)=a_{n}+\left(a_{n}^{2}+b_{n}^{2}-b_{n}\right) i \\
a_{n+1}=a_{n} \\
b_{n+1}=a_{n}^{2}+b_{n}^{2}-b_{n} \\
\Rightarrow\left\{\begin{array}{c}
a_{n}=\frac{\sqrt{3}}{2} \\
b_{n+1}=b_{n}^{2}-b_{n}+\frac{3}{4}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{c}
a_{n}=\frac{\sqrt{3}}{2} \\
b_{n+1}-\frac{1}{2}=\left(b_{n}-\frac{1}{2}\right)^{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{r}
a_{n}=\frac{\sqrt{3}}{2} \\
b_{n}-\frac{1}{2}=\left(-\frac{1}{2}\right)^{2^{n-1}}
\end{array}\right.
\end{array}
$$

Thus, $z_{2021}=\frac{\sqrt{3}}{2}+\left(\frac{1}{2}+\frac{1}{2^{2020}}\right) \mathrm{i}$.