$4 \cdot 59$ The equation $z^{6}+z^{3}+1=0$ has a complex root, on the complex plane this root has an argument $\theta$ between $90^{\circ}$ and $180^{\circ}$, find the degree measure of $\theta$.


[Solution] Let $\omega=z^{3}$. Then the original equation can be transformed into
$$
\begin{array}{l}
\omega^{2}+\omega+1=0, \\
\omega=\cos \frac{2 \pi}{3} \pm i \sin \frac{2 \pi}{3} .
\end{array}
$$

Since $z$ is the cube root of $w$, we have
$$
\begin{array}{l}
z_{1}=\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}, \\
z_{2}=\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}, \\
z_{3}=\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}, \\
z_{4}=\cos \frac{4 \pi}{9}+i \sin \frac{4 \pi}{9}, \\
z_{5}=\cos \frac{10 \pi}{9}+i \sin \frac{10 \pi}{9}, \\
z_{6}=\cos \frac{16 \pi}{9}+i \sin \frac{16 \pi}{9} .
\end{array}
$$

Among the arguments between $90^{\circ}$ and $180^{\circ}$, only $\frac{8 \pi}{9}$ is included, so $\theta=160^{\circ}$.