Calculate the Laurent development of the function $\displaystyle f(z)=\frac{1}{(z-a)(z-b)}$ in the region $\displaystyle \Omega: 0<|a|<|z|<|b|<+\infty$.

My attempt: I notice that f has 2 singularities, one in z=a, the other in z=b. By intuition, the radius of convergence of the series should be $\displaystyle \frac{|b|-|a|}{2}$ but it's not asked.

Ok so my problem is where do I center the Laurent series? There's the possibility that f is analytic in $\displaystyle \Omega$ (and so the Laurent series is worth the Taylor's series) but I'm unsure since it doesn't seem bounded when $\displaystyle z\to a^+$ and $\displaystyle z\to b^-$ so I'd bet it's not analytic in omega.

Now $\displaystyle \frac{1}{z-a}=\frac{1}{z} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n$ with $\displaystyle \left | \frac{a}{z} \right |<1 \Rightarrow |a|<|z|$ which is ok in the region omega. I could get a similar result for $\displaystyle \frac{1}{z-b}$ but it wouldn't be valid in omega so I don't really know how to go further.

So I can write $\displaystyle f(z)=\frac{1}{z(z-b)} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n$ and I'm stuck.