Example 3 The equation $z^{6}+z^{3}+1=0$ has a complex root, and on the complex plane, the argument of this root is between $90^{\circ}$ and $180^{\circ}$. Find the degree measure of $\theta$.
(2nd American Invitational Mathematics Examination)
From $\left(z^{3}-1\right)\left(z^{6}+z^{3}+1\right)=z^{9}-1$, we know that the complex roots satisfying the equation $z^{6}+z^{3}+1=0$ are the 9th primitive roots of unity, and the 9th roots of unity are $z_{1}=\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{2}, z_{2}=\cos \frac{4 \pi}{9}+i \sin \frac{4 \pi}{9}$,
$$
\begin{array}{l}
z_{4}=\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}, z_{5}=\cos \frac{10 \pi}{9}+i \sin \frac{10 \pi}{9}, z_{7}=\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}, z_{8}=\cos \frac{16 \pi}{9}+ \\
\text { isin } \frac{16 \pi}{9} \text {. }
\end{array}
$$

The argument that lies between $90^{\circ}$ and $180^{\circ}$ is only $\frac{8}{9} \pi=160^{\circ}$, which is the degree of the required $\theta$.