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$\def\zbar{\smash{\overline z}\vphantom z}$I would like to construct an approximation of the product \begin{equation} f(z) = \frac{1}{\zbar-a} \frac{1}{z-b}, \end{equation} where $a, b \in \mathbb{C}$, and $|{a}/{z}|, |{b}/{z}| <1$.

More precisely, I would like to obtain a separable expression of the form: \begin{equation} f(z) = \sum_{k=0}^\infty g_k(a,b) h_k(z, \overline{z}), \end{equation} where $g_k$ depends only on $a$ and $b$, and $h_k$ depends only on $z$ and $\overline{z}$. We can rewrite the product $f(z)$ as a double sum: \begin{equation} \frac{1}{\zbar-a} \frac{1}{z-b} = \frac{1}{|z|^2}\left(\sum_{j=0}^{\infty} \left(\frac{a}{\overline{z}}\right)^j \right) \left(\sum_{k=0}^\infty \left(\frac{b}{z}\right)^k \right) = \frac{1}{|z|^2} \sum_{j,\;k} \left(\frac{a}{\overline{z}}\right)^j \left(\frac{b}{z}\right)^k. \end{equation}

Unfortunately, the function $f$ is not analytic so we cannot use a Cauchy product to obtain the desired form.

Question: Is there a technique to obtain the desired factorization of $f$ when $|{a}/{z}|, |{b}/{z}| <1$?

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    $\begingroup$ Isn't that what you wanted, with $h_{jk}=\overline{z}^{-j-1}z^{-k-1}$, $g_{jk}=a^jb^k$ and if desired, you can label the $(j,k)$ by a single index. $\endgroup$ Sep 27, 2021 at 22:30
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    $\begingroup$ Cauchy product has to do with convergence, not with analyticity. $\endgroup$
    – GH from MO
    Sep 27, 2021 at 23:22
  • $\begingroup$ @Christian Remling, thank you for your answer. I am not sure that I understand your point. How do you define $g_k$ and $h_k$ to only have one infinite sum. $\endgroup$
    – Mathieu
    Sep 28, 2021 at 5:01
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    $\begingroup$ I understand @ChristianRemling to mean the following: Define, e.g., the sequence $(j,k)=(0,0),(0,1),(1,0),(0,2),(1,1),(2,0),(0,3),(1,2),(2,1),(3,0),(0,4),...$ and sum over the terms $g_{jk} h_{jk} $ according to that sequence ... $\endgroup$ Sep 28, 2021 at 6:55
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    $\begingroup$ Yes, what Michael explained is what I had in mind: use a bijection $\mathbb N\to\mathbb N^2$ to label the $g,h$ by a single index if desired. $\endgroup$ Sep 28, 2021 at 15:10

1 Answer 1

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Answer

I am interested in forming a factorization for $f(z) = ({\bar{z} -a})^{-1} \cdot ({z -b}^{-1})$ of the following form \begin{equation} f(z) = \sum_{k=0}^\infty g_k(a,b) h_k(z, \overline{z}), \end{equation} where $a, b \in \mathbb{C}$, $|{a}/{z}|, |{b}/{z}| <1$, $g_k$ depends only on $a$ and $b$, and $h_k$ depends only on $z$ and $\overline{z}$. We can rewrite the product $f(z)$ as a double sum: \begin{equation} \frac{1}{\overline{z}-a} \frac{1}{z-b} = \frac{1}{|z|^2}\left(\sum_{j=0}^{\infty} \left(\frac{a}{\overline{z}}\right)^j \right) \left(\sum_{k=0}^\infty \left(\frac{b}{z}\right)^k \right) = \frac{1}{|z|^2} \sum_{j,\;k} \left(\frac{a}{\overline{z}}\right)^j \left(\frac{b}{z}\right)^k. \end{equation}

To obtain the desired factorization for $f$, we use a bijection from $\mathbb{N}^2 \to \mathbb{N}$, such bijection exists because $\mathbb{N}^2$ is countable. For instance, we can use the Cantor pairing $\sigma$ (https://en.wikipedia.org/wiki/Pairing_function) given by: \begin{equation} \begin{aligned} \sigma \; \colon \; & \mathbb{N}^2 \to \mathbb{N},\\ & (j, k) \mapsto \sigma(j,k) = \frac{1}{2}(j+k)(j+k+1)+k. \end{aligned} \end{equation} The inverse of the Cantor pairing is denoted $\sigma^{-1}$. Let define $g_{j,k}(a,b) = a^j b^k$, and $h_{j,k}(z, \bar{z}) = \bar{z}^{-j-1} z^{-k-1}$ for $j, k\in \mathbb{N}$. The desired factorization is given by \begin{equation} f(z) = \sum_{l=0}^\infty g_{\sigma^{-1}(l)}(a,b)h_{\sigma^{-1}(l)}(z,\bar{z}). \end{equation}

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