Prove the roots of $z^7+7z^4+4z+1=0$ lie inside a circle of radius $2$.

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.

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.

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.)