6. Let the complex sequence $z_{n}$ satisfy for all positive integers $n$ that $z_{n} \cdot \overline{z_{n+1}}=z_{n^{2}+n+1}, z_{n} \cdot z_{n+2}=z_{n+1}^{2}-1$, given that $z_{2}=2+\mathrm{i}$, then the value of $z_{9}$ is $\qquad$ .
Let $n=1$ we get $\left\{\begin{array}{l}z_{1} \cdot \overline{z_{2}}=z_{3}, \\ z_{1} \cdot z_{3}=z_{2}^{2}-1\end{array} \Rightarrow\left\{\begin{array}{l}z_{1}(2-\mathrm{i})=z_{3}, \\ z_{1} \cdot z_{3}=2+4 \mathrm{i}\end{array} \Rightarrow\left\{\begin{array}{l}z_{1}=1+\mathrm{i}, \\ z_{3}=3+\mathrm{i}\end{array}\right.\right.\right.$
or $\left\{\begin{array}{l}z_{1}=-1-\mathrm{i}, \\ z_{3}=-3-\mathrm{i} .\end{array}\right.$ If $z_{1}=1+\mathrm{i}, z_{2}=2+\mathrm{i}, z_{3}=3+\mathrm{i}$, conjecture $z_{n}=n+\mathrm{i}$.
Assume the conclusion holds for $n \leqslant k$, then $z_{k+1}=\frac{z_{k}^{2}-1}{z_{k-1}}=\frac{(k+\mathrm{i})^{2}-1}{k-1+\mathrm{i}}$
$$
=\frac{\left(k^{2}-2+2 k \mathrm{i}\right)(k-1-\mathrm{i})}{(k-1)^{2}+1}=\frac{k^{3}-k^{2}+2+\left(k^{2}-2 k+2\right) \mathrm{i}}{k^{2}-2 k+2}=k+1+\mathrm{i} \text {, }
$$

i.e., the conclusion also holds for $n=k+1$. Thus, $z_{n}=n+\mathrm{i}$ holds, and $z_{9}=9+\mathrm{i}$;
If $z_{1}=-1-\mathrm{i}, z_{2}=2+\mathrm{i}, z_{3}=-3-\mathrm{i}$, then $z_{4}=\frac{z_{3}^{2}-1}{z_{2}}=4+\mathrm{i}$,
$$
z_{5}=\frac{z_{4}^{2}-1}{z_{3}}=-5-\mathrm{i}, \cdots, z_{9}=\frac{z_{8}^{2}-1}{z_{7}}=-9-\mathrm{i} \text {. In summary, } z_{9}= \pm(9+\mathrm{i}) \text {. }
$$