Consider a function $f(z, w)$ of two complex variables. The function is symmetric with respect to $z$ and $w$. When $\Re(z)>0$ and $\Re(w)>0$, the function is analytic in its two variables. When $\Re(w)>0$ and $\Re(z)\leq 0$, the function has poles in $z$ at positions $q\_i$. There are countably infinitely many poles that extend to $\Re(z) \rightarrow -\infty$. There may also be points of accumulation of poles. 


I would like to show that it is possible to holomorphically extend $f(z, w)$ to the region $\Re(z) \leq 0$ and $\Re(w) \leq 0$, except at the points in which $z=q\_i$ or $w=q\_j$.


I have been reading on holomorphic functions of several complex variables and it seems that if a holomorphic function $g(z, w)$ is defined in a connected subset $M$ of $\mathbb{C}^2$, then it can be uniquely holomorphically extended to the convex hull of $M$.


So, in my problem, it seems clear that $f(z, w)$ can be extended holomorphically into the region with $\Re(z) \leq 0$, $\Re(w) \leq 0$, $ Im(z) \neq 0$ and $Im(w) \neq 0$. By drawing some more pictures, I think I can show what I intended.


Can somebody more knowledgeable confirm this? My background is in physics, so perhaps this is a very naive question. Thanks.



Computing the convex hull of a region of $\mathbb{C}^2$
The result about extending to convex hull is not as general as what you state, although the erroneous generality seems irrelevant to your actual issue: it is for (connected) "tube domains", meaning sets described by conditions on the real parts only. (Further, some mild growth condition is necessary.)


A general key-word/phrase about extendability is "domain of holomorphy"...


In the tube domain case, and in $\mathbb C^2$, we can draw a not-toooo-misleading picture of the real parts: the "poles" then become vertical and horizontal lines $\Re(z)=q\_i$ (with $\Re(w)$ arbitrary) and vice-versa. 


In a fixed region $\{(z,w):\Re(z)\ge x\_o,\; \Re(w)\ge u\_o\}$ without accumulation points of poles, we can multiply through by a polynomial (depending on $x\_o, u\_o$) to reduce to the theorem on tube domains.


At accumulation points, I don't think there is a *general* recipe. Perhaps in the case that the dependence of the residue(s) on the *other* variable is very simple, a Weierstrass-Hadamard product expression (for $f(1/z)$...) could allow reduction to the cited result.


A particular worry would be that if there are "too many" poles near an accumulation point, the growth of the infinite product (as in Hadamard's theorem comparing "genus" and "order") might be too great to fit the theorem.


EDIT: to be a little more precise in an exaggeratedly simple case of an accumulation point $z\_o=0$ of *simple* (for example) poles at $z\_1,z\_2,\ldots \to 0$: the preliminary idea of the infinite product to consider is $\prod\_n (z-z\_n)$ but this doesn't converge. Second try: $F(z)=\prod\_n (1-{z\_n\over z})$. For $z\_n\to 0$ pretty strongly, this does converge for $z\not=0$. In general, as in Weierstrass-Hadamard, some exponential factors will be needed. Then $F(z)\cdot f(z,w)$ is holomorphic at $z=0$... so maybe the theorem applies, and this extends, and then we need to divide the extension by $F(z)$. This does introduce an essential singularity (in $z$) at $z=0$, yes, but we still do have some sort of extension.


