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.
 A: 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.
A: Use Rouché's theorem with the choice $f(z) = z^7 + 1$ and $g(z) = 7z^4 + 4z$.  Then you need to show that for $z = 2e^{i\theta}$; i.e., on the boundary of the disk of radius $2$ at the origin, $|g(z)| < |f(z)|$.  Then $f+g$ and $f$ have the same number of zeroes in the disk.  Then because $f$ has all seven zeroes on the unit circle, the result follows.
(I edited the previous version to make it a bit easier to show $g$ satisfies the condition.)
A: Assume that $z$ is a root of the equation with $|z| \ge 2$. Then
$$
 1 = \left|\frac{7z^4+4z+1}{z^7} \right|
\le \left|\frac{7}{z^3}\right| + \left|\frac{4}{z^6}\right| +\left| \frac{1}{z^7} 
\right| \\
\le \frac{7}{8} + \frac{4}{64} + \frac{1}{128} = \frac{121}{128} < 1
$$
gives a contradiction.
The idea is to show that if $|z| \ge 2$ then the modulus of $z^7$ is larger than the modulus of $7z^4+4z+1$, so that the sum of these terms can not be zero.
Remark: According to WolframAlpha, the equation $z^7 + 7z^4+4z+1 = 0$ has one solution $z \approx -1.86283$, so that the estimate $|z| < 2$ for all solutions is not bad.
