Let $A_z(R_1,R_2)$ denote the open annulus about $z$ with inner radius $R_1$ and outer radius $R_2$.
According to the basic theorem of Laurent series, if $f\in Hol(A_z(R_1,R_2))$ then $f$ has a laurent series expansion about $z$ which converges locally uniformly in the annulus.
But what about cases where $f$ is holomorphic in multiple, mutually disjoint annuli?
For instance, consider $$f=\frac{1}{(z-1)z}$$
It has singularities both at $z=0$ and $z=1$ and is holomorphic in both the anuulus $A_0(0,1)$ and $A_0(1,\infty)$.
Does that mean that $f$ can be represented as two different Laurent series about $0$? each converging in a different annulus? If so, what can we tell about the relation between their coefficients? And what can the series tell us about $f$?
If there is only a single Laurent expansion of $f$ about $0$, then where does it converge?
Perhaps most importantly though, do the answers to these questions depend on the specific example of $f$ at all?
Yes, functions can certainly have different expansions in different annuli. A particular expansion is only meaningful for the annulus in which it converges.
This is a pretty general result, for holomorphic functions with isolated singularities. If you have studied power series beforehand, you may recall that a power series expansion about $x_0$ has a radius of convergence equal to the distance from $x_0$ to the nearest singularity. A Laurent series of $f$ about $x_0$ behaves similarly, only now the inner radius (of the annulus in which it converges) is determined by the singularity furthest from $x_0$ which is closer to $x_0$ than the point of expansion.
For a specific example, $$g(z) = \frac{1}{(z - 1)(z^2 + 4)}$$ has singularities at $1, \pm 2i$ and is holomorphic everywhere else. It therefore has three Laurent series about $z = 0$: one converges in $A_0(0,1)$, another in $A_0(1,2)$, and the last in $A_0(2,\infty)$. Note how the radii of these annuli are determined by the distances between the singularities and the center of the expansion, which in this case is $0$.