11. (This question is worth 20 points) A complex number sequence $\left\{z_{n}\right\}$ is called "interesting" if $\left|z_{1}\right|=1$, and for any positive integer $n$, we have $4 z_{n+1}^{2}+2 z_{n} z_{n+1}+z_{n}^{2}=0$. Find the largest constant $C$ such that for all interesting sequences $\left\{z_{n}\right\}$ and any positive integer $m$, we have $\left|z_{1}+z_{2}+\cdots+z_{m}\right| \geq C$.
Solution: Consider the interesting complex sequence $\{z_n\}$. By induction, we know that $z_n \neq 0 \ (n \in \mathbf{N}^*)$. From the given condition, we have
$$
4\left(\frac{z_{n+1}}{z_n}\right)^2 + 2\left(\frac{z_{n+1}}{z_n}\right) + 1 = 0 \ (n \in \mathbf{N}^*),
$$

Solving this, we get $\frac{z_{n+1}}{z_n} = \frac{-1 \pm \sqrt{3} \mathrm{i}}{4} \ (n \in \mathbf{N}^*)$. Therefore, $\frac{|z_{n+1}|}{|z_n|} = \left|\frac{z_{n+1}}{z_n}\right| = \left|\frac{-1 + \sqrt{3} \mathrm{i}}{4}\right| = \frac{1}{2}$, hence
$$
|z_n| = |z_1| \cdot \frac{1}{2^{n-1}} = \frac{1}{2^{n-1}} \ (n \in \mathbf{N}^*).
$$
5 points
Furthermore, we have
$$
|z_n + z_{n+1}| = |z_n| \cdot \left|1 + \frac{z_{n+1}}{z_n}\right| = \frac{1}{2^{n-1}} \cdot \left|\frac{3 \pm \sqrt{3} \mathrm{i}}{4}\right| = \frac{\sqrt{3}}{2^n} \ (n \in \mathbf{N}^*).
$$

Let $T_m = |z_1 + z_2 + \cdots + z_m| \ (m \in \mathbf{N}^*)$.
When $m = 2s \ (s \in \mathbf{N}^*)$, using (2) we get
$$
T_m \geq |z_1 + z_2| - \sum_{k=2}^{s} |z_{2k-1} + z_{2k}| > \frac{\sqrt{3}}{2} - \sum_{k=2}^{\infty} |z_{2k-1} + z_{2k}| = \frac{\sqrt{3}}{2} - \sum_{k=2}^{\infty} \frac{\sqrt{3}}{2^{2k-1}} = \frac{\sqrt{3}}{3}.
$$

When $m = 2s + 1 \ (s \in \mathbf{N}^*)$, from (1) and (2) we know
$$
|z_{2s+1}| = \frac{1}{2^{2s}} \frac{\sqrt{3}}{2} - \sum_{k=2}^{\infty} |z_{2k-1} + z_{2k}| = \frac{\sqrt{3}}{3}.
$$

When $m = 1$, $T_1 = |z_1| = 1 > \frac{\sqrt{3}}{3}$.
The above shows that $C = \frac{\sqrt{3}}{3}$ satisfies the requirement. $\qquad$
On the other hand, when $z_1 = 1, z_{2k} = \frac{-1 + \sqrt{3} \mathrm{i}}{2^{2k}}, z_{2k+1} = \frac{-1 - \sqrt{3} \mathrm{i}}{2^{2k+1}} \ (k \in \mathbf{N}^*)$, it is easy to verify that $\{z_n\}$ is an interesting sequence. In this case,
$$
\begin{array}{r}
\lim_{x \to \infty} T_{2s+1} = \lim_{s \to \infty} \left|z_1 + \sum_{k=1}^{s} (z_{2k} + z_{2k+1})\right| \\
= \lim_{s \to \infty} \left|1 + \sum_{k=1}^{s} \frac{-3 - \sqrt{3} \mathrm{i}}{2^{2k+1}}\right| = \left|1 + \frac{-3 + \sqrt{3} \mathrm{i}}{8} \cdot \frac{4}{3}\right| = \frac{\sqrt{3}}{3},
\end{array}
$$

This shows that $C$ cannot be greater than $\frac{\sqrt{3}}{3}$.
In summary, the required $C$ is $\frac{\sqrt{3}}{3}$.
20 points