Show that a complex polynomial has a root inside a disk

Question

Here is an exercise from Guillemin and Pollack's text:

Let $$p(z)=z^m+a_1z^{m-1}+\dots+a_m$$ be a complex polynomial function. Prove that if $r^m> |a_1|r^{m-1}+\dots+|a_m|$, then $p$ has a root inside the disk of radius $r$, $\{|z|< r\}$.

I believe I need to use

Proposition. Let $W$ be a smooth region in $\mathbb C$ whose boundary contains no zeros of the polynomial $p$. Then the total number of zeros of $p$ inside $W$, counting multiplicities, is the degree of the map $p/|p|: \partial W\to S^1$.

In the context of the problem, $W=\{|z| < r\}$. Assume that $p$ has no zeros inside $W$. Then $\deg (p/|p|)=0$. The idea is to construct a homotopy from $z^m$ to $p/|p|$ and use that $\deg$ is homotopy invariant. I can construct a homotopy from $p(z)$ to $z^m$ by $p_t(z)=tp(z)+(1-t)z^m$, but how to construct the needed homotopy and how to use the condition on $r$?

(Solutions using other methods are also welcome.)

Answer 1

Obviously if we take $s_1,s_2,...,s_m$ as the roots of the polynomial we have$$p(z)=z^m+a_1z^{m-1}+\dots+a_m=(z-s_1)(z-s_2)\dots(z-s_m)$$which concludes $$|s_1s_2\dots s_m|=|a_m|$$assume there is no root inside $\{|z|<r\}$ i.e. $|s_1|,|s_2|,\cdots,|s_m|\ge r$. Therefore $$|a_m|=|s_1s_2\dots s_m|\ge r^m$$but according to $r^m> |a_1|r^{m-1}+\dots+|a_m|$ we have $r^m>|a_m|$ which is in contrary with $r^m\le |a_m|$. This proves that our assumption is wrong and at least one root is in $\{|z|<r\}$

Answer 2

Actually, Rouché's Theorem can be applied to $f(z)=z^m$ andd $g(z)=a_1z^{m-1}+...+a_m$ on the disk $|z|<r$ and yields that $f$ and $p=f+g$ have the same number of zeros in this disk, counting multiplicities. Therefore $p$ has even $m$ zeros in the disk, counting multiplicities.

Answer 3

Assume that $|z|\ge r$. Then as $~\dfrac{|z|}{r}\ge 1~$ so $~\dfrac{|z|^k}{r^k}\le\dfrac{|z|^{m-1}}{r^{m-1}}~$ for $k<m$ and thus \begin{align} |p(z)|&\ge |z|^m - |a_{m-1}z^{m-1}+\dots+a_1z+a_0|\\[.8em] &\ge |z|^m-\frac{|z|^{m-1}}{r^{m-1}}(|a_{m-1}|r^{m-1}+\dots+|a_1|r+|a_0|)\\[.8em] &>|z|^m-r|z|^{m-1}>0 \end{align} so that $p$ has no root outside the disk with radius $r$.