The roots of $z^4+z^3+1=0$ are in $\frac{3}{4}<|z|< \frac{3}{2}$. How can we show that all the roots of the complex polynomial
$$p(z)=z^4+z^3+1$$
lie in $\frac{3}{4}<|z|< \frac{3}{2}$? This is an exercise in complex analysis. 

Use Rouché's theorem.
Note that for $|z|=\frac{3}{2}$, we have $|(z^4+z^3+1)-z^4| = |z^3+1| \le 1+ |z|^3 = 1+ (\frac{3}{2})^3 < (\frac{3}{2})^4 = |z|^4 $, hence $z \mapsto z^4+z^3+1$ and $z \mapsto z^4$ have the same number of zeros in $B(0,\frac{3}{2})$ (that is, all four zeros)
Let $\epsilon>0$. 
Then for $|z| = \frac{3}{4}+\epsilon$, we have $|(z^4+z^3+1)-1| = |z^4+z^3| \le |z|^4+|z|^3 = 
(\frac{3}{4}+\epsilon)^3+ (\frac{3}{4}+\epsilon)^4$. Since $f(\epsilon ) = (\frac{3}{4}+\epsilon)^3+ (\frac{3}{4}+\epsilon)^4$ is continuous, and $f(0) <1$, it follows that for some $\epsilon>0$, we have $f(\epsilon)<1$.
Hence for $|z| = \frac{3}{4}+\epsilon$ we have $|(z^4+z^3+1)-1| < 1 = |1|$, hence $z \mapsto z^4+z^3+1$ and $z \mapsto 1$ have the same number of zeros in $B(0,\frac{3}{4}+\epsilon)$ (that is, there are no zeros in $B(0,\frac{3}{4}+\epsilon)$). 
It follows that all zeros are in $B(0,\frac{3}{2}) \setminus \overline{B}(0,\frac{3}{4})$.
