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I have this function of the complex variable $z$:

$f(z)=(1-z^{-1})^\alpha$

and I would like to do the following:

  1. Replace $1/z$ with $x$, thus $f(z)$ becomes $f(x)=(1-x)^{\alpha}$;
  2. Expand $f(x)$ using Taylor, about $x=0$.

Then I would get a polynomial expansion with positive powers of $x$, that would become negative powers of $z$. Can I do that? What are the mathematical implication? It is like I am treating a complex function as a real one. Is that correct? Can I then go back to the formulation in $z$?

Thanks!

Joe
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1 Answers1

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It corresponds to making a Laurent expansion (centered at zero) of the complex valued function $f(z)$ for $|z|>1$. Definitions depends upon the value of $a$. If e.g. $a$ is not an integer then in order to define $f$ you should make a cut from 1 to 0. (Like for $\sqrt{z}$ where you make a cut from 0 to e.g. $-\infty$).

H. H. Rugh
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  • Hi! Thanks for your answer! So, under which conditions a Laurent series of a function f(z) corresponds to the Taylor series of f(x), where x=1/z?

    Is it due to the fact that the function is holomorphic everywhere except in zero? If yes, can you give some references for this?

    Thanks a lot! :)

     – Joe Sep 10 '16 at 15:41
  • You may have a look in https://en.wikipedia.org/wiki/Laurent_series (In your case the function is analytic between $r=1$ and $R=+\infty$, the latter in fact included). But it doesn't cover quite your problem. I haven't seen this precise type of problem before – H. H. Rugh Sep 10 '16 at 15:53