9. Given the complex number $z$ satisfies $\left|2 z+\frac{1}{z}\right|=1$. Then the principal value range of the argument of $z$ is $\qquad$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
\text { Let } z=r(\cos \theta+\mathrm{i} \sin \theta) \Rightarrow 2 z+\frac{1}{z}=\left(2 r+\frac{1}{r}\right) \cos \theta+\mathrm{i}\left(2 r-\frac{1}{r}\right) \sin \theta \\
\Rightarrow\left(2 r+\frac{1}{r}\right)^{2} \cos ^{2} \theta+\left(2 r-\frac{1}{r}\right)^{2} \sin ^{2} \theta=1 \Rightarrow 4 \cos 2 \theta=1-\left(4 r^{2}+\frac{1}{r^{2}}\right) \leqslant-3 \\
\Rightarrow \cos 2 \theta=2 \cos ^{2} \theta-1 \leqslant-\frac{3}{4} \Rightarrow \cos ^{2} \theta \leqslant \frac{1}{8} \Rightarrow-\frac{\sqrt{2}}{4} \leqslant \cos \theta \leqslant \frac{\sqrt{2}}{4} \text {. } \\
\text { Therefore, } \arg z \in\left[\arccos \frac{\sqrt{2}}{4}, \pi-\arccos \frac{\sqrt{2}}{4}\right] \cup\left[\pi+\arccos \frac{\sqrt{2}}{4}, 2 \pi-\arccos \frac{\sqrt{2}}{4}\right] \text {. }
\end{array}
$$