9. (16 points) If the complex number $z$ satisfies $4 z^{-2011}-3 \mathrm{i} z^{-2010}-3 \mathrm{i} z^{-1}-4=0$, find the range of $t=\overline{\left(\frac{3-4 \mathrm{i}}{z}\right)}+\overline{(3+4 \mathrm{i}) z}$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
2,9. Transform the known equation into
$$
\begin{array}{l}
4 z^{2011}+3 \mathrm{i} z^{2010}+3 \mathrm{i} z-4=0 \\
\Rightarrow z^{2010}(4 z+3 \mathrm{i})=4-3 \mathrm{i} z \\
\Rightarrow|z|^{2010}|4 z+3 \mathrm{i}|=|4-3 \mathrm{i} z| .
\end{array}
$$

Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$. Then
$$
\begin{array}{l}
|4 z+3 \mathrm{i}|^{2}-|4-3 \mathrm{i} z|^{2} \\
=|4 a+(4 b+3) \mathrm{i}|^{2}-|(4+3 b)-3 a \mathrm{i}|^{2} \\
=16 a^{2}+(4 b+3)^{2}-(4+3 b)^{2}-9 a^{2} \\
=7\left(a^{2}+b^{2}-1\right)=7\left(|z|^{2}-1\right) .
\end{array}
$$

If $|z||4-3 \mathrm{i} z|$, then
$$

In equation (2), we have $|4 z+3 \mathrm{i}|1$, which also leads to a contradiction.
Therefore, $|z|=1$.
Let $z=\mathrm{e}^{\mathrm{i} \theta}, 3+4 \mathrm{i}=5 \mathrm{e}^{\mathrm{i} \alpha}$.
Then $t=(3+4 \mathrm{i}) z+(3-4 \mathrm{i}) \bar{z}$
$$
\begin{array}{l}
=5 \mathrm{e}^{\mathrm{i}(\theta+\alpha)}+5 \mathrm{e}^{-\mathrm{i}(\theta+\alpha)} \\
=10 \cos (\theta+\alpha) \in[-10,10] .
\end{array}
$$

Thus, the required range is $[-10,10]$.