# 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 $$\mathcal{Z}\{x[n]\} = X(z)$$.

• Are you not including the regions $\{\lvert z\rvert<z_0\}$ and $\{\lvert z\rvert>z_3\}$? Nov 10, 2021 at 3:55
• @Sandejo Certainly that can happen. I'm not sure if in those cases $f(z)$ necessary have a pole at $z=0$ or $z=\infty$. Nov 10, 2021 at 3:59
• @Benjamin Yes, I mean the poles of $f(z)$. Nov 13, 2021 at 17:37
• There exists a Laurent series in each annulus, and no annulus in which a Laurent series exists can contain a pole. That is what you already said, what are your doubts about it? Nov 15, 2021 at 16:00
• @MartinR I couldn't prove the following statement formally: "For every ring between two poles there is a unique Laurent series and there is no other Laurent series for $f(z)$." Also I don't know what happens if there are infinitely many poles like $\frac{1}{1 - \cos(z)}$ or if we have poles at infinity. Nov 15, 2021 at 16:27

$$f$$ 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 $$A_k$$.
• Great answer, thanks. Nov 16, 2021 at 4:46