Expansions of Functions Analytic at Infinity

1. Mar 20, 2014

ferret123

1. The problem statement, all variables and given/known data

Prove that if f(z) is analytic at infinity, then it has expansion of the form

$f(z) = \sum_{n=0}^{+\infty} \frac{a_{n}}{z^{n}}$

converging outside some disk.

2. The attempt at a solution

I know that for f(z) to be analytic at infinity we want to consider the composite function g(w) = f(\frac{1}{w}), then if g is analytic at 0 f will be analytic at infinity.

I've been attempting to Taylor expand g(w) around the point w=0 and then substituting the appropriate f terms in. However I'm not sure if this is a valid method as I can't see much pattern in the derivatives.

So far I have

$g(w) = g(0) + g'(0)w + \frac{g''(0)w^{2}}{2} + \frac{g'''(0)w^{3}}{3!} + ...$

as well as

$g'(w) = \frac{-f'(\frac{1}{w})}{z^{2}}$
$g''(w) = \frac{2f'(\frac{1}{w})}{w^{3}} + \frac{f''(\frac{1}{w})}{w^{4}}$
$g'''(w) = \frac{-6f'(\frac{1}{w})}{w^{4}} - \frac{6f''(\frac{1}{w})}{w^{5}} - \frac{f'''(\frac{1}{w})}{w^{6}}$

meaning i'm just gaining more of each derivative each time and I'm not sure if this is the right way to go about this. Any help or tips would be much appreciated.

Last edited: Mar 20, 2014
2. Mar 20, 2014

pasmith

What you need here is that if $f$ is analytic at infinity then $g$ is analytic at zero.

You don't need to do that. It is enough to note that if $g$ is analytic at zero then it has a power series expansion
$$g(w) = \sum_{n=0}^\infty a_n w^n$$
which converges inside some disc.

3. Mar 20, 2014

ferret123

Ok I've really over complicated it then, thanks! So then when I change back to $f$ I get the desired result analytic outside of a disk with radius 1/radius of the $g$ disk.

The next part of the question is looking for an expansion of this form for for $\frac{z-1}{z+1} \& \frac{z^{2}}{z^{2} - 1}$. I think for the first of these I can re-express as $1 - \frac{2}{z+1}$ and then expand this?