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:enter image description here 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 $\mathcal{Z}\{x[n]\} = X(z)$.

  • $\begingroup$ Are you not including the regions $\{\lvert z\rvert<z_0\}$ and $\{\lvert z\rvert>z_3\}$? $\endgroup$
    – Sandejo
    Nov 10, 2021 at 3:55
  • $\begingroup$ @Sandejo Certainly that can happen. I'm not sure if in those cases $f(z)$ necessary have a pole at $z=0$ or $z=\infty$. $\endgroup$
    – S.H.W
    Nov 10, 2021 at 3:59
  • $\begingroup$ @Benjamin Yes, I mean the poles of $f(z)$. $\endgroup$
    – S.H.W
    Nov 13, 2021 at 17:37
  • $\begingroup$ 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? $\endgroup$
    – Martin R
    Nov 15, 2021 at 16:00
  • $\begingroup$ @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. $\endgroup$
    – S.H.W
    Nov 15, 2021 at 16:27

1 Answer 1


$f$ 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 $A_k$.
  • $\begingroup$ Great answer, thanks. $\endgroup$
    – S.H.W
    Nov 16, 2021 at 4:46

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