I have a confusion regarding finding the residues of singularities from the Laurent series, for example consider the function
$$f(z) = \frac{1}{(z+1)(z+2)(z+3)}$$
and I have to find the residue at say $z = -1$, First we expand the Laurent series at the singularity $(z = 1)$, but then there would be $3$ series each for the regions:
$$|z + 1| < 2 \\ 2 < |z + 1| < 3 \ and \\ |z+1| > 3$$
So which one should I use to find the residue and why ?
Thanks for any help !
The singularities of a meromorphic function are isolated. That implies about any pole $\zeta$ of $f$ I can find a small annulus which does not contain any of the other poles of $f$, so we can "ignore them" and find the Laurent series about $\zeta$ without caring about these other poles.
Here, about $z=-1$, I can choose to not care about $z=-2,z=-3$. I can choose the annulus: $$\mathcal{A}=\left\{z\in\Bbb C:\frac{1}{2}>|z+1|>\frac{1}{4}\right\}$$
And this doesn't contain $-2,-3$. For $z\in\mathcal{A}$ it is true that: $$f(z)=\frac{1}{2}\frac{1}{z+1}\frac{1}{1+(z+1)}\frac{1}{1+\frac{1}{2}(z+1)}\\=\frac{1}{2}(z+1)^{-1}\left(\sum_{n\ge0}(-1)^n(z+1)^n\right)\left(\sum_{n\ge0}\frac{(-1)^n}{2^n}(z+1)^n\right)$$
Which, when expanded, yields a Laurent series about $\zeta:=-1$ valid in an annulus $\mathcal{A}$ centred at $\zeta=-1$ for which you can find the explicit coefficients. The residue is especially easy to find since it is just $\frac{1}{2}$ the constant coefficient of the product of series on the right; the constant term there is $(-1)^0(-1)^02^{-0}=1$, so the residue is simply $\frac{1}{2}$.
Laurent series expansions, like Taylor series expansions, are local. "(the residue of a) Laurent series about the pole $\zeta$ valid in an annulus centred at $\zeta$" is all you need, by definition. The region $|z+1|>3$, say, is completely irrelevant.