8. (1997 National High School Competition Question) 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
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8. Fill $k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4} \leqslant \theta \leqslant k \pi+\frac{\pi}{2}+\frac{1}{2} \arccos \frac{3}{4}(k=0,1)$. Reason: Let $z=r(\cos \theta+\mathrm{i} \sin \theta)$, then $\left|2 z+\frac{1}{2}\right|^{2}=\left(4 r^{2}+\frac{1}{r^{2}}\right)+4 \cos 2 \theta=1,4 r^{4}+(4 \cos 2 \theta-1) r^{2}+1=0$.

This equation holds if and only if the quadratic equation $4 x^{2}+(4 \cos 2 \theta-1) x+1=0$ has two positive roots $x_{1} 、 x_{2}$ $\Leftrightarrow$ the discriminant $\Delta=$ $(4 \cos 2 \theta-1)^{2}-4^{2} \geqslant 0$ and $x_{1}+x_{2}=-\frac{1}{4}(4 \cos 2 \theta-1)>0, x_{1}, x_{2}=\frac{1}{4}>0 \Leftrightarrow \cos 2 \theta-\frac{3}{4} \Leftrightarrow 2 k \pi+\pi-$ $\arccos \frac{3}{4} \leqslant 2 \theta \leqslant 2 k \pi+\pi+\arccos \frac{3}{4} \Leftrightarrow k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4} \leqslant \theta \leqslant k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4}(k=0,1)$.