Unit circle and roots Consider the polynomial $( f(x) = x^7 - 4x^3 + x + 1 )$. All the zeroes of the above polynomial are plotted in the complex plane i.e., the Argand plane. How many are within a unit distance from the origin?
Details and Assumptions:
The repeated roots are counted with multiplicity i.e.,$ ( (2x - 1)^2 = 0 )$ has two solutions within a unit distance from origin and not one.
There is a very neat and simple way of doing without W|A. Please refrain from using any such mathematical computational software.
 A: Let $g(x) = -4x^3$, then if $|x| = 1$ we have
$|f(x)-g(x)|= |x^7+x+1| \le 3 < 4 = |-4 x^3| = |g(x)|$, hence Rouché's theorem tells us that $f$ and $g$ have the same number of zeros inside
the unit circle.
A: As a numerically closer alternative to the other answer:
The dominant binomials of the polynomial $f(x)$ are $x^7-4x^3$ for the larger and $-4x^3+1$ for the smaller roots. Indeed the product of the reduced binomials is
$$
g(x)=(x^4-4)(x^3-\tfrac14)=x^7-\tfrac14x^4-4x^3+1 = f(x)-\tfrac14x^4-x
$$

Numerically, the root sets compare as
\begin{array}{rl|l}
&\text{numerical roots of }g(x)  &  \text{numerical roots of }f(x)\\ \hline
+\sqrt2&=+1.414214  &+1.313850\\
-\sqrt2&=-1.414214  &-1.401144\\
\pm i\sqrt2&=\pm i·1.414214  &+0.024884\pm i·1.456126\\
\tfrac12\sqrt[3]2&=+0.629961  &+0.792062\\
\tfrac14\sqrt[3]2(-1\pm\sqrt3)&=-0.314980\pm i·0.545562  &-0.377268\pm i·0.425477\\
\end{array}

By Rouché, on $|x|=1$, $|x^4-4|\ge 4-1=3$, $|x^3-\frac14|\ge \frac34$ and thus
$$
|f(x)-g(x)|\le \tfrac54< \tfrac94=3·\tfrac34\le |g(x)|
$$
which implies again the same number of roots inside and outside the unit circle for $f$ and $g$.
