Complex analysis proof about $|f(z)|$ I have to prove the following and have absolutely no idea where to start:
If $f$ is holomorphic in $|z|>R$ and its limit at $\infty$ is $0$, then $\exists \; m \in \mathbb{N}$ such that $|f(z)| \leq \frac{c}{|z|^m}$ for $|z|>R$ and some $c>0$.
 A: Michael M gave a wondeful hint above. 
Define the function $g:U\to \mathbb{C}$, where $U$ consists of $z$ where $|z| < \frac{1}{R}$, and $g(z) = f(z^{-1})$, with $g(0) = 0$. Check that this function is holomorphic. Here is a hint for this step. Prove that $g$ on the open set $U\setminus \{0\}$ is holomorphic and the limit of $f(z)$ as $z\to 0$ exists (so $0$ is a removable singularity).
Since $0$ is a zero of $g$, it follows that $g(z) = z^mg_1(z)$ where $g_1(z)\not = 0$ (assuming the function we are initially starting with $f$, is not identically zero, but then the problem is trivial). Can you take it from here? 
A: $g(z) = f(1/z)$ for $z\neq 0$
$g$ is now holomorphic for $0<|z|<R$.
$g$ admits a Laurent expansion around $0$.
Furthermore $\lim_{z\rightarrow 0}g(z) = 0$
So $g(z) = \sum_{n=1}^{\infty}c_n z^n$
or $f(z) = g(1/z) = \sum_{n=1}^{\infty}\frac{c_n}{z^n}$.
since $\lim_{z\rightarrow\infty} zf(z) = c_1$, by definition for every $c\in \mathbb{R}^+$ there exists $R$ such that for $|z|>R$ $|zf(z)|<c \Leftrightarrow |f(z)|<\frac{c}{|z|}$
So $m=1$ always works.
