$f$ holomorphic in an annulus $A(R) = \{ z \in \mathbb{C} : |z| > R\}$ Question. Suppose that $f$ holomorphic in an annulus $A(R) = \{ z \in \mathbb{C} : |z| > R\}$ and define $g(z) = f(1/z)$. Clearly $g$ has an isolated singularity at $0$, the question now asked us to:
(a) Show that the singularity of $g$ at $0$ is a pole if and only if given $R' > R$, there are constants $C$ and $N$ such that $|f(z)| \leq C|z|^N$ for all $z \in A(R')$.
(b) Show that if $|f(z)| \leq C|z|^N$ for all $z \in A(R)$, then there is a polynomial $p$ such that $f-p$ is bounded in $A(R)$.
My attempt.
I know that the singularity of $g$ is a pole if and only if its Laurent series has finitely many negative powers. So I write the LS for $g(z)$ in $|z| < \frac{1}{R}$.
Suppose that the Laurent series for $f$ in $A(R)$ is $f(z) = \sum_{n = -\infty}^{\infty}c_nz^{n}$. Then
$$g(z) = f(1/z) = \sum_{n = -\infty}^{\infty}c_nz^{-n}$$
For this to have finitely many negative powers, we want the Laurent series for $f$ to have finitely many positive powers, which means $|f(z)| \leq C|z|^N$ for some constants $C$ and $N$.
But I'm not sure why do we need $R' > R$ here? It would be really appreciated if someone could confirm if I'm on the right track and point me to the right direction.
Edit. With help from the answers I was able to finish my first proof. Now I show my attempt at part b of the question:
Suppose that $|f(z)| \leq C|z|^N$ for all $z \in A(R)$. We write the Laurent Series for $f$ as above. Let $z \to \infty$. Then $f(z)$ behaves like the highest positive power term. For this to be bounded by $C|z|^N$ we require $f$ to have finitely many positive powers.
Let $M$ to be the highest power (I'd imagine $M$ has to be smaller than $N$ but I'm not sure?), that is,
$$f(z) = \sum_{n = -\infty}^{M}c_nz^{n}$$
Take $p(z) = \sum_{n = 0}^{M}c_nz^{n}$, then $f-p$ consists only of negative powers and so $(f-p)(z) \to 0$ as $z \to \infty$. This means $f-p$ is bounded.
Is this correct?
 A: Why do you need $R'>R$? Because you may have a singularity on the circle $\{z:z=R\}$. For example, take $f(z)=z+1/(z-1)$ and $R=1$.
This has a simple pole at $\infty$. But any estimate $|f(z)|\le C|z|^N$
breaks down when we let $z\to 1$, so cannot be valid in the set
$\{z:|z|>1\}$.
In effect the proof boils down to the fact that
a function $g$ holomorphic in a deleted
neighbourhood of $0$ has a pole there iff $|z^ng(z)|$ is bounded near zero,
and that reduces to the characterisation of removable singularities.
A: You are on the right track. Asserting that $g$ has a pole at $0$ means that, for some $N\in\mathbb N$, $c_n=0$ if $n>N$. But, if this occurs, then\begin{align}g(z)&=\sum_{n=-\infty}^Nc_nz^{-n}\\&=z^{-N}\sum_{n=-\infty}^Nc_nz^{N-n}\\&=z^{-N}\sum_{n=0}^\infty c_{N-n}z^n\end{align}Let $h(z)=\sum_{n=0}^\infty c_{N-n}z^n$, which is defined when $\lvert z\rvert<\frac1R$. If $R'>R$, let $M=\max_{\lvert z\rvert\leqslant1/R'}\bigl\lvert h(z)\bigr\rvert$. Then$$\lvert z\rvert\leqslant\frac1{R'}\implies\bigl\lvert g(z)\bigr\rvert\leqslant\lvert z\rvert^{-N}M.$$Can you take it from here?
