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

The equation $z^{6}+z^{3}+1=0$ has a complex root, and on the complex plane, the argument $\theta$ of this root lies 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 $\theta=160^{\circ}$.