8. The complex sequence $z_{0}, z_{1}, \cdots$ satisfies
$$
\left|z_{0}\right|=1, z_{n+1}=\frac{i z_{n}}{\overline{z_{n}}} \text {. }
$$

If $z_{2011}=1$, then $z_{0}$ can have $\qquad$ possible values.
8. $2^{2011}$.

Obviously, for any non-negative integer $n$, we have
$$
\left|z_{n}\right|=1 \text {. }
$$

Let $z_{n}=\mathrm{e}^{\mathrm{i} \theta_{n}}\left(\theta_{0} \in[0,2 \pi)\right)$. Then
$$
\begin{array}{l}
\mathrm{e}^{\mathrm{i} \theta_{n+1}}=\frac{\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{x}}{2}+\theta_{n}\right)}}{\mathrm{e}^{-\mathrm{i} \theta_{n}}} \\
\Rightarrow \theta_{n+1}=2 \theta_{n}+\frac{\pi}{2} \\
\Rightarrow \theta_{n+1}+\frac{\pi}{2}=2\left(\theta_{n}+\frac{\pi}{2}\right)=\cdots=2^{n}\left(\theta_{0}+\frac{\pi}{2}\right) .
\end{array}
$$

From $z_{2011}=1$, we get
$$
\theta_{2011}=2 k \pi(k \in \mathbf{Z}) \text {, }
$$

i.e., $2^{2011}\left(\theta_{0}+\frac{\pi}{2}\right)=2 k \pi+\frac{\pi}{2}$.
By $\theta_{0} \in[0,2 \pi)$, we get
$$
\begin{array}{l}
2^{2010} \pi \leqslant 2 k \pi+\frac{\pi}{2}<2^{2010} \times 5 \pi \\
\Rightarrow \frac{2^{2011}-1}{4} \leqslant k<\frac{5 \times 2^{2011}-1}{4} \\
\Rightarrow 2^{2009} \leqslant k<5 \times 2^{2009} .
\end{array}
$$
- Therefore, the number of $z_{0}$ that satisfy the condition is
$$
5 \times 2^{2009}-2^{2009}=2^{2011} \text { (individuals). }
$$