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Setting: Let $f$ be meromorphic on $\mathbb{\hat{C}} = \mathbb{C} \cup \{\infty\}$. Let $\{p_i\}$ be the $k$ number of poles of $f$. Let $n_i$ denote the orders of each of the $p_i$.

Question: Why is it that

$$ g(z) = f(z) \cdot (z - p_1)^{n_1} \cdot \ldots \cdot (z - p_k)^{n_k} $$

is an analytic function on $\mathbb{C}$? It seems to me that that given $f(z)$ is still a factor of $g(z)$, that $g(p_i)$ is still going to be undefined for all $i \in \{1, \ldots , k\}$.

Note: The motivation for this post is in trying to understand a statement made in this discussion.

Community
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user1770201
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2 Answers2

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The singularities of meromorphic functions can be multiplied away as you stated. After this, you can close the holes by assigning $g(p_1)$ the value of $\lim_{z \to p_1} f(z)(z-p_1)^{n_1}$. This would not necessarily work for essential singularities.

In particular the Laurent series of a meromorphic function $f$ expanded about a pole of order $n_1$ is given by:

$$f(z) = a_{-n_1}(z-p_1)^{-n_1} + a_{-n_1 + 1}(z-p_1)^{-n_1+1} + \cdots + a_{-1}(z-p_1)^{-1} + a_0 + a_1 (z-p_1) + a_2 (z-p_1)^2 + \cdots$$

Then multiplying by $(z-p_1)^{n_1}$ cancels the pole.

Then $$g(z) = (z-p_1)^{n_1} f(z) = a_{-n_1} + a_{-n_1 + 1}(z-p_1) + a_{-n_1+2}(z-p_1)^2 + \cdots$$

Joel
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    Another way to think about this, is to think about how to define the function $g(x)=f(x)\cdot x$ when $f(x) = 1/x$. I think any reasonable person would say $g(0)=1$. Of course, you can't arrive at this by explicit evaluation of $f(0)$, which does not exist, but it is convention to declare $g(x) \equiv 1$ – Joel Apr 28 '14 at 16:59
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    More precisely, the limit $\lim_{z\rightarrow p_i}g(z)$ is well defined as one still cannot just plug in $f(p_i)$. – Alex R. Apr 28 '14 at 16:59
  • So then, just to be clear, what we really have is that $g(z) = \prod(z-p_i)^{n_i}f(z)$ is analytic in $\mathbb{C}$ where, strictly speaking, we have that $g(p_i)$ is defined as $\lim_{z \to p_i} \prod(z-p_i)^{n_i}f(z)$. Is this the case? – user1770201 Apr 28 '14 at 17:34
  • Yes that is the case. – Joel Apr 28 '14 at 17:48
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If $f$ had a pole $z_1$ with order $a$. Then $g(z)=f(z)(z-z_1)^a$ is continuous in $z_1$ and holomorphic in a region of $z_1$. You can see then that it is holomorphic to this region .

Haha
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