# If $f$ analytic in $|z|>1$ and $|f(z)|<|z|^n$, then $f$ has finitely many zeros in $|z|>2$.

Let $f(z)$ be analytic in $\Omega = \{|z|>1\}$. Suppose that $f$ satisfies $|f(z)| < |z|^n$ for all $z \in \Omega$ and for some n> 0. Prove that either $f$ has finitely many zeros in $\{|z|>2\}$ or $f$ is identically zero.

My argument was :

$\Omega$ is biholomorphically equivalent to $U = \{0<|z|<1\}$ via the map $z\rightarrow 1/z$. So we can look at the equivalent problem : - $f$ is analytic in $U$ and satisfies $|f(z)|<|z|^n$. Then if $f$ has infinitely many zeros in $|z|<1/2$, the zeros will accumulate on either $|z|=1/2$ or at zero. If they accumulate on $|z|=1/2$ then $f$ is identically zero by identity principle. Again, $|f|<1$ for all $z \in U$. Hence $f$ is bounded and holomorphic in the punctured disk and thus has a removable singularity at the origin. Thus zeros cannot accumulate at the origin unless $f$ is identically zero.

• How did you use the bound by $|z|^n?$ Jan 10, 2014 at 1:11
The method is $1-\epsilon$ correct. However, the function to look at is $z^n f(1/z),$ which is bounded at $0$ by assumption.
• Here $f(1/z)$ is the composition of $f$ with the biholomorphism that I mentioned. If I understand correctly then - the equivalent problem would need to be slightly modified as $|f(z)|< 1/|z|^n$ for all $z \in U$. This still makes $f$ bounded, however not by 1 but by some other constant. Am I correct? Jan 10, 2014 at 1:38
• @SouravD it is bounded once you multiply it by $z^d,$ which is done in my answer (notice that the original function is not bounded at infinity, so neither will the transformed one be at zero). Jan 10, 2014 at 1:46
• I only need the transformed function to be bounded in a punctured neighborhood of zero like $U = \{0<|z|<1\}$. This it indeed is. From here Riemann Theorem about removable singularities tells us that it is bounded at zero as well. Jan 10, 2014 at 1:57