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Here's the definition.

Let $X$ be a Riemann surface and $Y$ be an open subset of $X$. A meromorphic function on $Y$ is a holomorphic function $f\colon Y' \rightarrow \mathbb{C}$ satisfying the following conditions, where $Y'\subset Y$ is a open subset.

(1) Every point $p\in Y - Y'$ is an isolated point.

(2) For every point $p\in Y - Y'$, $\lim_{x\rightarrow p} |f(x)| = \infty$.

The points of $Y - Y'$ are called the poles of $f$.

Then he stated in a remark:

Let $(U, z)$ be a coordinate neighborhood of a pole $p$ of $f$ with $z(p) = 0$. Then $f$ can be expanded in a Laurent series $f = \sum_{n = -k}^{\infty} c_n z^n$ in a neighborhood of $p$.

Why is this so? In other words, why $p$ cannot be an essential singularity?

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Maybe I'm confused about the order in which Forster makes definitions, but he seems to have defined poles in the above. Most people would then go on to define an essential singularity to be one that is neither removable nor a pole. So it seems by definition to me that it can't be an essential singularity because it is a pole. –  Matt Sep 20 '12 at 22:02
    
An essential singularity wouldn't satisfy $(2)$, because in the case of an essential singularity, the limit will not be well defined (by the Picard theorem). –  Ben A. Sep 20 '12 at 22:08
    
@BenA. Could we avoid using the Picard theorem? –  Makoto Kato Sep 20 '12 at 22:13
    
@Makoto. I'm still thinking about how to avoid it. But you see, that $(2)$ is the sticking point, right? –  Ben A. Sep 20 '12 at 22:17
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@Makoto: You can get away with something a lot weaker and more elementary than Picard (e.g., the Casorati-Weierstrass theorem will suffice). –  Micah Sep 20 '12 at 22:37
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1 Answer

up vote 3 down vote accepted

As pointed out in the commentaries, assumption $(2)$ ensures that the singularity is a pole.

The statement is local, so let's say $f$ is a holomorphic function on $U\setminus{0}$, where $U\subset\mathbb{C}$ is a neighborhood of the origin in the complex plane, and that $\lim_{z\to 0}{|f(z)|} = \infty$. This allows us to assume, that $f$ doesn't vanish on $U\setminus{0}$ (by possibly shrinking $U$). Thus $g = 1/f$ is holomorphic on $U\setminus{0}$ and bounded near zero. Therefore is has a unique holomorphic extension to $U$, obviously given by $g(0) = 0$ and this zero point is isolated, hence it has a finite order of vanishing. But this order of vanishing is, by definition, the order of the pole of $1/g = f$, i.e. $f$ has a pole of finite order in $0$ (and especially not an essential singularity).

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