Prove the roots of $z^7+7z^4+4z+1=0$ lie inside a circle of radius $2$. Problem: Prove that all the roots of the equation $z^7+7z^4+4z+1=0$ lie inside the circle of radius $2$ centered at the origin.
This question came in the roots of unity section of a summer course. My attempts have been to factor and hope something nice pops out, which ended up in the following: $$z(z^3+2)^2+3z^4+1=0$$ $$\frac{z^8-1}{z-1}+3z^4=(z^4+z)(z^2+z-3).$$ They all look horrendous. Other thoughts would be to use some inequality to prove $|z|<2$ (am-gm, cauchy, etc. seem to fail me). Not entirely sure what to do from here.

This question can be solved using Roché's Theorem. Consider the holomorphic function
\begin{align}
h(z) := f(z) + g(z) = z^7 + 7z^4 + 4z + 1
\end{align}
with $f(z) = z^7$ and $g(z) = 7z^4+4z+1$. Consider $K := \{z \in \mathbb{C}: |z| < 2\}$. Then $\partial K$ is a circle of radius $2$ centered at the origin. On $\partial K$, that means for $|z| = 2$, we have
\begin{align}
|g(z)| = |7z^4+4z+1| \leq 7|z|^4 + 4|z| + 1 = 121 < 128 = |z|^7 = |f(z)|.
\end{align}
Then, by Rouché's Theorem, we have that $f$ and $f+g$ have the same number of zeros inside $K$. Since $f$ has all his seven zeros inside $K$ so does $f+g$ and we are done.
