How many roots have modulus less than $1$?

Question

If roots of the equation
$$x^7 - 4x^3 + x + 1=0$$

are plotted on the Argand plane, how many of them have distance from the origin less than $1$?

I found, by plotting the rough curve of $y= x^7 - 4x^3 + x + 1$ that it has three real roots, out of which two of them have modulus greater than $1$ and one has modulus less than $1$. But I don't know how to do the same for non - real roots. The answer given in my book is that $3$ roots have modulus less than $1$.

Answer 1

Rouches theorem: let $f,g$ be holomorphic in open set $U$ and $C$ boundary in $U$, of a disc inside $U$. If $|f|>|g|$ for all $z$ on the circle $C$ then $f$ and $f+g$ have same number of zeros inside $C$ (counted with multiplicity).

Let $g(z)=z^7+z+1$, and $f(z)=-4z^3$. Both are clearly holomorphic functions.

Then for $|z|=1$ i.e. $z$ on the unit circle (since it is in your question), we have $$|f(z)|=|-4\cdot 1|=4 \mbox{ and } |g(z)|\leq |z|^7+|z|+1=3.$$ Thus, $|g(z)|<|f(z)|$ for all $z$ inside unit circle $C$.

Can you complete the solution now?

Answer 2

Along the unit circle $|x| = 1$, the third degree term dominates the rest. Therefore, there are three zeroes inside the circle (counting multiplicity).

The idea is that compared to the polynomial $-4x^3$ (which has $3$ zeroes), the rest (namely $x^7 + x + 1$) is not large enough to "push" any of the zeroes out of the unit circle, and it's not large enough to "push" any zeroes from the outside and in.

The exact statement is called Rouché's theorem. It says:

If $f$ and $g$ are holomorphic inside and on some closed contour $K$, and on $K$ the inequality $|f(x)| > |g(x)|$, then $f$ and $f+g$ have the same number of zeroes inside $K$.

In this case, $f(x) = -4x^3$ and $g(x) = x^7 + x + 1$, with $K$ being the unit circle.