Polynomial $1+3z^m+5z^n$ ($1<m<n$) - Annulus $\frac{1}{3}< |z| < 1$

Question

Rouché's theorem : Let $D \subset \mathbb{C}$ a domaine and $f,g: D \to \mathbb{C}$ two holomorphic functions in $D$. Let $C$ a closed path contained in the interior of $D$. If $|f(z)+g(z)| < |f(z)|+|g(z)|$, $\forall z \in \mathbb{C}$, then $f$ and $g$ have the same number of zeroes in the interior of $C$.

Question : Show that all zeroes of the polynomial $1+3z^m+5z^n$ ($1<m<n$) are located in the annulus $\frac{1}{3}< |z| < 1$.

I think I have to use Rouché's theorem, but it is unclear how to use it. Is anyone could help me to solve this problem?

Thanks!

Answer 1

Problems like this are almost always manufactured, so there is usually a strong hint in the formulation.

For example, you want to show that there are $n$ zeroes inside $|z|=1$, and since $z \mapsto z^n$ has $n$ zeros inside the unit disk, this suggests that $z \mapsto 5 z^n$ would be a good candidate. So try $g(z) = - 5 z^n$.

$|1+3 z^m| \le 4 < 5 = |g(z)|$ for $|z| = 1$.

For another example, you want to show that there are no zeros inside $|z|= {1 \over 3}$, and since $z \mapsto 1$ has no zeros there, this suggests that $g(z)=-1$ would be a good candidate.

In the latter case, you should note that $n > m \ge 2$ when performing the estimate.

$|3z^m + 5 z^n| \le 3 {1\over 3^m}+ 5 {1 \over 3^n}< 3 {1\over 3^2}+ 5 {1 \over 3^2} ={8 \over 9} < 1 = |g(z)|$ for $|z| = {1 \over 3}$.