11. (20 points) A sequence of complex numbers $\left\{z_{n}\right\}$ is called "interesting" if $\left|z_{1}\right|=1$, and for any positive integer $n$, we have
$$
4 z_{n+1}^{2}+2 z_{n} z_{n+1}+z_{n}^{2}=0 \text {. }
$$

Find the largest constant $C$ such that for all interesting sequences $\left\{z_{n}\right\}$ and any positive integer $m$, we have
$$
\left|z_{1}+z_{2}+\cdots+z_{m}\right| \geqslant C \text {. }
$$
11. The largest constant $C$ is $\frac{\sqrt{3}}{3}$.

Consider the interesting complex sequence $\left\{z_{n}\right\}$.
By induction, we know $z_{n} \neq 0\left(n \in \mathbf{Z}_{+}\right)$.
From the condition, we have
$$
\begin{array}{l}
4\left(\frac{z_{n+1}}{z_{n}}\right)^{2}+2\left(\frac{z_{n+1}}{z_{n}}\right)+1=0\left(n \in \mathbf{Z}_{+}\right) \\
\Rightarrow \frac{z_{n+1}}{z_{n}}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{4}\left(n \in \mathbf{Z}_{+}\right) \\
\Rightarrow \frac{\left|z_{n+1}\right|}{\left|z_{n}\right|}=\left|\frac{z_{n+1}}{z_{n}}\right|=\left|\frac{-1 \pm \sqrt{3} \mathrm{i}}{4}\right|=\frac{1}{2} \\
\Rightarrow\left|z_{n}\right|=\left|z_{1}\right| \frac{1}{2^{n-1}}=\frac{1}{2^{n-1}}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$

Thus, $\left|z_{n}+z_{n+1}\right|=\left|z_{n}\right|\left|1+\frac{z_{n+1}}{z_{n}}\right|$
$$
=\frac{1}{2^{n-1}}\left|\frac{3 \pm \sqrt{3} i}{4}\right|=\frac{\sqrt{3}}{2^{n}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Let $T_{m}=\left|z_{1}+z_{2}+\cdots+z_{m}\right|\left(m \in \mathbf{Z}_{+}\right)$.
When $m=2 s\left(s \in \mathbf{Z}_{+}\right)$, using equation (2), we get
$$
\begin{array}{l}
T_{m} \geqslant\left|z_{1}+z_{2}\right|-\sum_{k=2}^{s}\left|z_{2 k-1}+z_{2 k}\right| \\
>\frac{\sqrt{3}}{2}-\sum_{k=2}^{\infty}\left|z_{2 k-1}+z_{2 k}\right| \\
=\frac{\sqrt{3}}{2}-\sum_{k=2}^{\infty} \frac{\sqrt{3}}{2^{2 k-1}}=\frac{\sqrt{3}}{3} .
\end{array}
$$

When $m=2 s+1\left(s \in \mathbf{Z}_{+}\right)$, from equations (1) and (2), we have
$$
\begin{array}{l}
\left|z_{2 s+1}\right|=\frac{1}{2^{2 s}}\frac{\sqrt{3}}{2}-\sum_{k=2}^{\infty}\left|z_{2 k-1}+z_{2 k}\right|=\frac{\sqrt{3}}{3} \text {. }
\end{array}
$$

When $m=1$, $T_{1}=\left|z_{1}\right|=1>\frac{\sqrt{3}}{3}$.
The above shows that $C=\frac{\sqrt{3}}{3}$ meets the requirement.
Furthermore, when $z_{1}=1, z_{2 k}=\frac{-1+\sqrt{3} \mathrm{i}}{2^{2 k}}$,
$$
z_{2 k+1}=\frac{-1-\sqrt{3} \mathrm{i}}{2^{2 k+1}}\left(k \in \mathbf{Z}_{+}\right)
$$

it is easy to verify that $\left\{z_{n}\right\}$ is an interesting sequence. At this time,
$$
\begin{array}{l}
\lim _{s \rightarrow \infty} T_{2 s+1}=\lim _{s \rightarrow \infty}\left|z_{1}+\sum_{k=1}^{s}\left(z_{2 k}+z_{2 k+1}\right)\right| \\
=\lim _{s \rightarrow \infty}\left|1+\sum_{k=1}^{s} \frac{-3-\sqrt{3} \mathrm{i}}{2^{2 k+1}}\right| \\
=\left|1+\frac{-3+\sqrt{3} \mathrm{i}}{8} \cdot \frac{4}{3}\right|=\frac{\sqrt{3}}{3},
\end{array}
$$

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