Approximation of the product $(\bar{z} - a)^{-1} \cdot (z - b)^{-1}$

Question

$\def\zbar{\smash{\overline z}\vphantom z}$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 $|{a}/{z}|, |{b}/{z}| <1$?

Answer 1

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}