Find all the possible laurent expansions of $f(z)=\frac{1}{\sqrt{1-z^2}}$

Question

I'm trying to find all the possible laurent expansions of $f(z)=\frac{1}{\sqrt{1-z^2}}$. In this case, there're two poles of the function at $\pm1$, but I'm not pretty sure how many cases I need to consider here? Is there only one case? Also, I'm told that I may use that the nth derivative of $\frac{1}{\sqrt{1-x^2}}$ at $x=0$ is $\frac{(2n-1)!!}{2^n}$, how can I use this identity? Thanks a lot for the help:)

Answer 1

$(1-z^2)^{-\frac 12}$

By the binomial theorem we get:

$(1-z^2)^{-\frac 12} = 1^{-\frac 12} + (-\frac 12) 1^{-\frac 32} (-z^2) + (-\frac 12)(-\frac 32)(\frac 12) (1^{-\frac 52})(-z^2)^2 +\cdots $

$1 +\frac 12 z^2 + \frac 38 z^4 + \frac 5{16}z^6 + \cdots$ And this will converge when $|z| < 1$

What can we do to find a series that will converge outside this range?

$(1-z^2)^{-\frac 12} = -iz^{-1}(1-z^{-2})^{-\frac 12}$

And do a similar expansion

Answer 2

1. $f(z)$ is a multivalued function, thus you should first clarify on which branch you will expand this function;
2. There are four cases for expanding $f(z)$ in power series, but only one is Laurent series, that is expansion at infinity.
3. At $z=0$ the expansion is Taylor series, at $z=\pm1$ -- Puiseux series.