For $n \in \mathbb{N}_+$, let $z_n = a_n + b_n i$ ($a_n, b_n \in \mathbb{R}$). Then
$$\begin{aligned} a_{n+1} + b_{n+1}i &= z_{n+1} = \overline{z_n}(1 + z_n i) \\ &= \overline{z_n} + |z_n|^2 \cdot i \\ &= a_n - b_n i + (a_n^2 + b_n^2)i, \end{aligned}$$
and hence $a_{n+1} = a_n$, $b_{n+1} = a_n^2 + b_n^2 - b_n$.
And by $z_1 = \frac{\sqrt{3}}{2}$ we know that $a_1 = \frac{\sqrt{3}}{2}$, $b_1 = 0$, so $a_n = \frac{\sqrt{3}}{2}$, and thus
$$b_{n+1} = b_n^2 - b_n + \frac{3}{4},$$
namely, $b_{n+1} - \frac{1}{2} = b_n^2 - b_n + \frac{1}{4} = \left(b_n - \frac{1}{2}\right)^2$.
Hence, when $n \ge 2$,
$$\begin{aligned} b_n &= \frac{1}{2} + \left(b_1 - \frac{1}{2}\right)^{2^{n-1}} \\ &= \frac{1}{2} + \left(-\frac{1}{2}\right)^{2^{n-1}} \\ &= \frac{1}{2} + \frac{1}{2^{2^{n-1}}}. \end{aligned}$$
Consequently,
$$$$
$z_{2021} = a_{2021} + b_{2021}i = \frac{\sqrt{3}}{2} + \left(\frac{1}{2} + \frac{1}{2^{2^{2020}}}\right)$ i.