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I'm not studying several complex variables, but I may need to use some elementary results from the subject—particularly those regarding singularities.

I know Hartogs' Extensions Theorem boils down to: "isolated singularities of holomorphic functions of multiple variables are removable", but I'm having a difficult time wrapping my head around it.

One example that does make sense to me is $\frac{1}{z-w}$; there, the singularities are indeed non-isolated. However, what about something like $\frac{1}{z-a}\frac{1}{w-b}$ or $\frac{z-a}{w-b}$?—where $a$ and $b$ are (possibly non-distinct) complex numbers. Are the singularities of these functions truly removable? If so, how do you go about analytically continuing the functions to take values at those singularities—i.e., what are the "correct" values there? On the other hand, if they aren't removable, how do they not satisfy the hypotheses of Hartogs' Extension Theorem?

MCS
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    They aren't isolated. – Angina Seng Jun 21 '19 at 21:05
  • Ah. I see. I wasn't thinking of it like that. Thanks. :) – MCS Jun 21 '19 at 21:10
  • @MCS A few things to note regarding Hartogs' extension theorem: It does not hold for $\mathbb{C}$, it tells you more than what you mention above. It tells you that a function which is holomorphic on $G \backslash K$, where $G$ is a domain in $\mathbb{C}^{n >1}$ and $K \subset G$ is a compact set with $G \backslash K$ connected, extends to a holomorphic function on $G$. That is, there exists a holomorphic function $\tilde{f} \in \mathcal{O}(G)$ such that $\tilde{f} \vert_{G\backslash K} \equiv f$. – AmorFati Jul 09 '19 at 01:24

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