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ferret123

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## Homework Statement

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

[itex]f(z) = \sum_{n=0}^{+\infty} \frac{a_{n}}{z^{n}}[/itex]

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

[itex]g(w) = g(0) + g'(0)w + \frac{g''(0)w^{2}}{2} + \frac{g'''(0)w^{3}}{3!} + ...[/itex]

as well as

[itex]g'(w) = \frac{-f'(\frac{1}{w})}{z^{2}}[/itex]

[itex]g''(w) = \frac{2f'(\frac{1}{w})}{w^{3}} + \frac{f''(\frac{1}{w})}{w^{4}}[/itex]

[itex]g'''(w) = \frac{-6f'(\frac{1}{w})}{w^{4}} - \frac{6f''(\frac{1}{w})}{w^{5}} - \frac{f'''(\frac{1}{w})}{w^{6}}[/itex]

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.

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