As for your more general question, here is what I think. Lets talk about your function f(z). It is a quotient of polynomials where the numerator is a constant and the denominator is a product of two linear factors. There is a theorem somewhere saying that such functions are analytic everywhere except at their singularities. In your case z = a or z = b. From this, you can see the f(z) must be analytic in \(\displaystyle \Omega\), which is an annulus.

Now, I'm not 100% sure on this, but talking about radius of convergence for a Laurent series can be a bit of a misnomer. We should really be talking about a region in which the series converges. For similar reasons, the whole issue of where you center the series is a bit superfluous as well, because of the following:

If you center the Laurent series around 0, it'll equal the Taylor series for f(z) since it is analytic on |z| < |a|, and it'll terminate at the nearest singularity, which is a.

Similarly, in omega, you can choose a point and derive a Taylor series about the point, and it'll terminate at whichever singularity is closer a or b. This will depend on the placement of said point in relation to a and b.

Now, the point of the Laurent series in this case would be to find a series representation of f(z) which will be analytic in the whole annulus omega. So, in this case we are not really looking at a point at which we want to center the series. Rather, we are looking at a way to manipulate f(z) to look like a the sum of a geometric series. In this example, it should be pretty easy, because the function is already looking ready.

So you have the first step, which is to split apart f(z). You can actually do without the splitting, but the downside is that you will have to multiply series and this can be messy. So:

\(\displaystyle \frac{1}{(z-a) (z-b)}=\frac{1}{(b-a) (z-b)}+\frac{1}{(a-b) (z-a)} = \frac{1}{b-a}(\frac{1}{z-b}-\frac{1}{z-a})\)

Now you have the following inequalities: |a| < |z| and |z| < |b|. You can see where this is going. From these we need to find expressions of z that have magnitude < 1 to use converging geometric series. From the first inequality we get |a\z| < 1 and from the second |z\b| < 1. Now you are pretty much ready.

\(\displaystyle \frac{1}{z-b} = -\frac{1}{b}\frac{1}{1-z/b} \) and \(\displaystyle \frac{1}{z-a} = \frac{1}{z}\frac{1}{1-a/z} \)

So now \(\displaystyle f(z) = \frac{1}{a-b}(\frac{1}{b}\frac{1}{1-z/b} +\frac{1}{z}\frac{1}{1-a/z})\). From here you can just use geometric series and maybe simplify a bit.