$$
\begin{array}{r}
\text { 5. Given a sequence } z_{0}, z_{1}, \cdots, z_{n}, \cdots \text { satisfying } z_{0}=0, z_{1} \\
=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2,
\end{array}
$$
5. Given a sequence of complex numbers $z_{0}, z_{1}, \cdots, z_{n}, \cdots$ satisfying $z_{0}=0, z_{1}$ $=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2$, $\cdots$. The number of $z_{n}$ contained within the circle $\mid z \mid=10$ is
5.5.

Since $z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right)=\alpha^{2}\left(z_{n-1}-z_{n-2}\right)$ $=\cdots=\alpha^{n}\left(z_{1}-z_{0}\right)=\alpha^{n}$, therefore,
$$
z_{n}-z_{n-1}=\alpha^{n-1}, \cdots, z_{1}-z_{0}=\alpha^{0}=1 \text {. }
$$

Adding the above $n$ equations, we get
$$
z_{n}=\alpha^{n-1}+\cdots+\alpha+1=\frac{\alpha^{n}-1}{\alpha-1} \text {. }
$$

Since $\alpha=1+\sqrt{3} i=2\left(\cos \frac{\pi}{3}+\sin \frac{\pi}{3}\right)$,
we have, from $\left|z_{n}\right|<10$,
$$
\left|\frac{2^{n}\left(\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}\right)-1}{\sqrt{3} \mathrm{i}}\right|<10,
$$

which means $\left|2^{n} \cos \frac{n \pi}{3}-1+\mathrm{i} 2^{n} \sin \frac{n \pi}{3}\right|<10 \sqrt{3}$.
Thus, $2^{2 n}-2^{n+1} \cos \frac{n \pi}{3}+1<300$.
Noting that $\left(2^{n}-1\right)^{2}=2^{2 n}-2^{n+1}+1 \leqslant 2^{2 n}-2^{n+1}$ $\cdot \cos \frac{n \pi}{3}+1 \leqslant 300$, we can conclude that $n \leqslant 4$, meaning there are 5 points $z_{n}$ inside the circle $|z|<10$, namely $z_{0}, z_{1}, z_{2}, z_{3}$, $z_{4}$.