Taylor (or Laurent) Expansion of $f(z)=(1-z^{-1})^\alpha$

Question

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!

Answer 1

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. $-\infty$).