I'm learning about Laurent Series and I'm having trouble with a certain part of the class notes.
The Laurent Series for a function $f(z)$ around a point $z_0$ is defined as
$$f(z) = \sum_{n=-\infty}^{+\infty} a_n (z-z_0)^n$$
Where
$$a_n = \frac{1}{2\pi i} \oint_C \frac{f(z')dz'}{(z' - z_0)}$$
I understand how this result is arrived at. However I have some questions:
I'm a little unnerved by the idea of summing from negative infinity to positive infinity. I get how you do that for integrals, but this is a discrete sum. In particular, can I rearrange it like this?
$$f(z) = a_0 + \sum_{n=1}^{+\infty} \big(a_n(z-z_0)^n + \frac{a_{-n}}{(z' - z_0)^n}\big)$$
I vaguely remember that rearranging the order of the terms in an infinite sum is a delicate business, and I want to know if this is allowed?
The text defines what the different kinds of poles and singularities mean, in terms of how many of the negative coefficients are zero. e.g. All $a_{-n} = 0$ from $n=-\infty$ to $n=-k$ means the pole is of order $k$. So far, so good. Next it says something I find quite cryptic:
It follows from these that poles are isolated singularities: $f(z)$ is only non-analytic at and immediately around $z_0$. This is easily proved, by multiplying through by $(z-z_0)^n$ [emphasis mine], which produces a series without negative powers, and thus represents an analytic function. If the function $(z-z_0)^n f(z)$ is analytic, $f(z)$ must be analytic except at $z = z_0$.
By the look of it, the bolded part seems to mean:
$$f(z)(z-z_0)^n = \sum_{n=-\infty}^{+\infty} a_n (z-z_0)^{2n}$$
If the previous rearranging is allowed, then it might mean:
$$f(z)(z-z_0)^n = a_0(z-z_0)^n + \sum_{n=1}^{+\infty} \big(a_n(z-z_0)^{2n} + a_{-n}\big)$$
Both of which I think are nonsense. You can't multiply the whole equation by something that depends on a sum's index. The LHS now has an $n$ in it, which doesn't make any sense.
How am I meant to interpret "multiplying through by $(z-z_0)^n$", and how exactly does it get rid of the negative powers?