Inequality for a complex root of a polynomial 
Given that $z$ is a root of $4z^4 + z+ 1 = 0$, show that $|z| \leq 1$. 

My attempt:
Suppose, for contradiction, that $|z| > 1$.
Consider the circle inequality on $4z^4 - (-z)$.
$$|4z^4 - (-z)| \geq ||4z^4|-|z|| = |4|z|^4 - |z|| > 3. \qquad \tag{$\star$}.$$
From our given information,
$$4z^4 + z = -1.$$
Hence, we know that
$$|4z^4 + z| = 1.$$
So $(\star)$ contradicts this, hence $|z| \leq 1$.  
Is this correct, if so, is there another way to approach this?
 A: Your proof is correct (replace "circle inequality" with "triangle inequality"). 
This is another approach. Let $r=2^{-1/4}<1$. For $|z|=r$, $$4r^4=|4z^4|=2>r+1\geq  |z+1|.$$ 
Hence by Rouche's theorem, $f(z)=4z^4$ and $g(z)=4z^4+z+1$  have the same number of roots (counting multiplicity) in $|z|<r$. Since $f$ has $4$ roots at $0$ and $g$ has degree $4$, it follows that $g$ has all its four complex roots in $|z|<r$ (stronger than $|z| \leq 1$ since $2^{-1/4}\approx 0.841$ ).
A: Your proof is correct, but I would elaborate a bit on the last part
of $(\star)$ since you estimate the difference of numbers.
For example like this:
$$
 1 = \lvert 4 z^4 + z \rvert \ge \lvert 4 z^4 \rvert - \lvert z \rvert
 = (4  \lvert z \rvert^3 - 1)\lvert z \rvert
$$
Now the assumption $\lvert z \rvert \ge 1$ leads to the contradiction 
$ 1 \ge 3$ since both factors on the RHS are positive.
In fact one can conclude that $\lvert z \rvert \le r$ must hold
where $r$ is the (unique) positive solution of
$$
 4 r^4 - r - 1 = 0
$$
which is $r \approx 0.821$.
