11. (20 points) Let the complex number $z$ satisfy $|z|=1$. Find the minimum possible value of 
$$
2021|1+z|+|1+z^{2}|+\cdots+|1+z^{4043}|
$$
11. First prove the general conclusion.

Notice,
$$
\begin{array}{l}
n|1+z|+\left|1+z^{2}\right|+\cdots+\left|1+z^{2 n+1}\right| \\
=\sum_{k=1}^{n}\left(|1+z|+\left|1+z^{2 k+1}\right|\right)+\sum_{k=1}^{n}\left|1+z^{2 k}\right| \\
\geqslant \sum_{k=1}^{n}\left|z-z^{2 k+1}\right|+\sum_{k=1}^{n}\left|1+z^{2 k}\right| \\
=\sum_{k=1}^{n}\left(|z|\left|1-z^{2 k}\right|+\left|1+z^{2 k}\right|\right) \\
=\sum_{k=1}^{n}\left(\left|1-z^{2 k}\right|+\left|1+z^{2 k}\right|\right) \\
\geqslant \sum_{k=1}^{n}\left|1-z^{2 k}+1+z^{2 k}\right|=2 n .
\end{array}
$$

When $z=-1$, the equality holds in the above expression.
Therefore, the minimum value of the original problem is 4042.