4. 2 .

Solution: According to the problem, when $n \geqslant 1$, $a_{n+1}^{2}-a_{n} \cdot a_{n+1}+a_{n}^{2}=$ $0, a_{n-1}^{2}-a_{n} \cdot a_{n-1}+a_{n}^{2}=0$, and $a_{n+1} \neq a_{n-1}$, so $a_{n-1}$, $a_{n+1}$ are the two roots of the quadratic equation $x^{2}-a_{n} \cdot x+a_{n}^{2}=0$,

Therefore, $a_{n-1}+a_{n+1}=a_{n}$, thus $a_{n+3}=a_{n+2}-a_{n+1}=$ $\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n}, a_{n+6}=-a_{n+3}=a_{n}$, hence,
$$
\begin{array}{l}
\sum_{n=0}^{2006} a_{n}=\sum_{n=0}^{2006}\left(a_{n+1}-a_{n+2}\right)=a_{1}-a_{2008}=a_{1}-a_{4}=a_{1} \\
-\left(-a_{1}\right)=2 a_{1}=2 .
\end{array}
$$
4. Given the complex sequence $\left\{a_{n}\right\}$ satisfies: $a_{n+1}^{2}-a_{n} a_{n+1}+$ $a_{n}^{2}=0(n \geqslant 0)$, and $a_{n+1} \neq a_{n-1}(n \geqslant 1), a_{1}=1$, then $\sum_{n=0}^{2006} a_{n}$ is $\qquad$.