An Application of Rouché's Theorem for the estimating the position of zeros

Question

I am working on the following exercise:

Let $p$ be a polynomial function of the form $p(z) = z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0$. Show that all zeros of $p$ are in the open circular disc $\Delta_r(0)$ with $r:= 1+ \max\{|a_{n-1}|, \dots, |a_{0}|\}$.

We can assume that $r>1$, otherwise the result is obvious. My approach so far was to provide an upper limit for $|p(z)|$ at the border of $\Delta_r(0)$, using the triangle inequality: \begin{equation} |z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0| < r^n + r \cdot r^{n-1} + \dots r \cdot r + r \end{equation} \begin{equation} = r^n + r \cdot \frac{r^n - 1}{r-1} < r^n + r^{n+1} \end{equation} By Rouché's Theorem, it follows that $q(z):= (r+1)z^n$ and $(p+q)(z)$ have the same number of zeros in $\Delta_r(0)$. However, I am not sure how to get from this result to the number of zeros for $p$ in $\Delta_r(0)$. Am I overlooking something?

Answer 1

Note that if $|z|=r$, then\begin{align}|a_{n-1}z^{n-1}+\cdots+a_1z+a_0|&\leqslant|a_{n-1}|r^{n-1}+\cdots+|a_1|r+|a_0|\\&\leqslant(r-1)(r^{n-1}+\cdots+r+1)\\&=r^n-1\\&<r^n\\&=|z|^n.\end{align}Therefore, by Rouché's theorem, $z^n$ and $p(z)$ have the same number of zeros in that region. But $z^n$ has $n$ zeros there (counting with multiplicities).

Answer 2

If $|z| = r$, then, using that $|a_i| \le r-1$ for each $i$, $$|a_{n-1}z^{n-1}+\dots+a_1z+a_0| \le (r-1)r^{n-1}+(r-1)r^{n-2}+\dots+(r-1)r+(r-1) = r^n-1 < |z^n|$$ So by Rouché, we see that the number of zeroes of $p$ in $\Delta_r(0)$ is the same as the number of zeroes as $z^n$ in $\Delta_r(0)$, namely $n$.