Laurent series of a meromorphic function Let $f(z)$ be a meromorphic function with the poles $\{z_0, z_1, \dots \}$. Suppose that we want to find the Laurent series representation of $f(z)$ centered at $z=0$. Can we claim that the only possible choices for the region of convergence is the rings between poles? And also there is a unique Laurent series for each of these rings? (i.e., $\{z_0\lt |z| \lt z_1, \ z_1\lt|z|\lt z_2, \dots \}$).
In short, for every ring between two poles there is a unique Laurent series and there is no other Laurent series for $f(z)$.
This is what I mean by the rings:
Intuitively, this seems correct to me because the Laurent series converges in the rings and this region cannot contains poles but I couldn't prove the claim formally. The main motivation of the question is the ROC(Region of convergence) of $\mathcal{Z}$-transform, which is defined as: $$\mathcal{Z}\{x[n]\} = X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n} , \ \ \text{ROC}\{X(z)\} = \{z \in \mathbb{C}:|X(z)|\lt \infty \}$$
Given $X(z)$, the goal is to find sequences such that $\mathcal{Z}\{x[n]\} = X(z)$.
 A: $f$ is homomorphic in $D = \Bbb C \setminus \{ z_1, z_2, z_3, \ldots \}$, where $(z_k)$ is a (finite or infinite) sequence of complex numbers without accumulation point in $\Bbb C$, and $f$ has a pole at each $z_k$.
We can assume that the $z_k$ are sorted with respect to increasing modulus:
$$
 0 \le |z_1| \le |z_2| \le |z_3| \le \ldots
$$
In the case of infinite poles we necessarily have $\lim_{k \to \infty} z_k = \infty$.
If $|z_k| < |z_{k+1}|$ for some $k$ then we define $A_k$ as the annulus
$$
 A_k = \{ z : |z_k| < |z| < |z_{k+1}| \} \, .
$$
In the case of finitely many poles $z_1, \ldots, z_N$ we additionally define
$$
 A_N = \{ z : |z_N| < |z|\} \, .
$$
The function $f$ is holomorphic in the each annulus $A_k$
and therefore can be developed into a Laurent series
$$ \tag{1}
 f(z) = \sum_{n=-\infty}^\infty a_n^{(k)} z^n \quad \text{for } z \in A_k \, .
$$
Now assume conversely that $f$ can be developed into a Laurent series
$$ \tag{2}
 f(z) = \sum_{n=-\infty}^\infty b_n z^n 
$$
which converges in some annulus $B = \{ z : r < |z| < R \}$. Then $f$ is holomorphic in $B$, i.e. it does not contain any of the poles. Let
$$
 m = \max \{ k : |z_k| \le r \} \, .
$$
There are two possible cases: $f$ has finitely many poles and $m = N$, or $|z_m| \le r < R \le |z_{m+1}|$. In both cases is $B \subseteq A_m$. Then $(1)$ with $k=m$ is a Laurent series for $f$ in $B$ and since Laurent series are unique, we necessarily have
$$
 b_n = a_n^{(m)} \quad \text{for } n=0, 1, 2, \ldots \, .
$$
This shows that

*

*$f$ has a Laurent series expansion in each annulus $A_k$ determined by two “consecutive” poles with different modulus (and, in the case of finitely many poles, a Laurent series expansion in the exterior $A_N$ of a disk), and

*Any Laurent series expansion of $f$ in an annulus with center at the origin is the restriction of one of these expansions in some $A_k$.

