1. If $z$ is a complex number, and $(z+1)(\bar{z}+1)=2004$, then the range of $|z|$ is $\qquad$ .
2.1. $[\sqrt{2004}-1, \sqrt{2004}+1]$.
Let $z=x+y \mathrm{i}, r=\sqrt{x^{2}+y^{2}}$, then $(x+1+y \mathrm{i})(x+1-y \mathrm{i})=2004$,
which means $(x+1)^{2}+y^{2}=2004$.
Thus, $r^{2}=2003-2 x$.
Also, $2004-(x+1)^{2}=y^{2} \geqslant 0$, so
$-\sqrt{2004}-1 \leqslant x \leqslant \sqrt{2004}-1$.
Therefore, $(\sqrt{2004}-1)^{2} \leqslant r^{2} \leqslant(\sqrt{2004}+1)^{2}$.
Hence, $\sqrt{2004}-1 \leqslant r \leqslant \sqrt{2004}+1$.