Is there a concept of meromorphic function of several complex variables? Is there a Taylor series description of such a function? More specifically, I am wondering if such several variable meromorphic functions should have poles only at isolated points, or they can have larger sets as poles? As an example, consider functions of two complex variables $z_1, z_2$. Is $f(z_1, z_2) = \frac{1}{z_1 - a_1}, a_1 \in \mathbb{C}$ a meromorphic function? What about the function$g(z_1, z_2) = \frac{1}{(z_1 - a_1)(z_2 - a_2)}, a_2 \in \mathbb{C}$? Thanks in advance!


A function $f$ of several complex variables is called meromorphic on $\Omega$ if for every $a \in \Omega$, there is a neighbourhood $U \ni a$ and holomorphic functions $p$ and $q$, where $q$ is not identically $0$, on $U$ such that $$ f(z) = \frac{p(z)}{q(z)} \qquad \text{for } z \in U \setminus q^{-1}(0). $$ In particular, the quotient of two holomorphic functions on $\Omega$ is meromorphic, so both your examples are meromorphic.

Note some differences with the one-variable case:

  • The singularity set of a meromorphic function is never discrete or even compact. In fact, it's an analytic variety of codimension $1$. (Hartogs' extension theorem shows that if $f$ is holomorphic on $\Omega \setminus K$ where $K$ is a compact subset of $\Omega$, then $f$ admits a unique extension to a holomorphic function on $\Omega$.)
  • We can't "extend" $f$ to its singularity set by pretending that $f=\infty$ there. For example look at $f(z,w) = z/w$ and see what happens as we approach the origin along various paths.
  • $\begingroup$ One quick question if I may (I am trying to better understand the singularity set of the meromorphic function). Suppose $S \subset \mathbb{C}^n$ is a closed set whose complement is unbounded. Can I construct a meromorphic function whose singularity set avoids $S$ and the function is bounded on $S$ (in dimension $1$, such things are easily achievable)? $\endgroup$ – user346998 Oct 28 '16 at 8:47
  • $\begingroup$ Not necessarily. For a simple counterexample, take $$S = \bigcup_{j=1}^\infty \{ z : |z| = j \}.$$ The singularity set of any meromorphic function (that is not entire) will intersect $S$. If $S$ is also compact, then yes: $1/(\text{suitably chosen first degree polynomial})$ will do. $\endgroup$ – mrf Oct 28 '16 at 8:52
  • $\begingroup$ Do you require that $K$ has codimension $\geq 2$? $\endgroup$ – punctured dusk Jul 30 '18 at 12:12

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