6. For a positive integer $n \geqslant 2007$, a complex number $z$ satisfies
$$
(a+1) z^{n+1}+a \text { i } z^{n}+a \text { i } z-(a+1)=0 \text {, }
$$

where the real number $a>-\frac{1}{2}$. Then the value of $|z|$ is $\qquad$ .
6.1.

Given $z^{n}[(a+1) z+a \mathrm{i}]=a+1-a \mathrm{i} z$, then $|z|^{n}|(a+1) z+a \mathrm{i}|=|a+1-a \mathrm{i} z|$. Let $z=x+y \mathrm{i}$, where $x, y$ are real numbers. Then
$$
\begin{array}{l}
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2} \\
=(2 a+1)\left(|z|^{2}-1\right) .
\end{array}
$$

If $|z|>1$, since $a>-\frac{1}{2}$, then
$$
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2}>0 \text {. }
$$

Thus, $|z|<1$, which is a contradiction.
Similarly, if $|z|<1$, it also leads to a contradiction.
Therefore, $|z|=1$.