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I am working through exercises in Lang's Complex Analysis 3e, and have a problem..

Chapter 5, Section 3, Problem 1: Show that the following series define a meromorphic function on $\mathbb{C}$ and determine the set of poles, and their orders.

(a) $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n!(n+z)}~~~~~~$(b) $\displaystyle\sum_{n=1}^{\infty}\frac{\sin(nz)}{n!(n^2+z^2)}~~~~~~$(c) $\dfrac{1}{z}+\displaystyle\sum_{\substack{n\neq0\\n=-\infty}}^{\infty}\Bigg[\frac{1}{z-n}+\dfrac{1}{n}\Bigg]$

Now each term $f_n$ of (a) is holomorphic everywhere except at $z=-n$ which is a simple pole. These will carry over to be simple poles in the sum, at all of the negative integers. Similarly, (b) has simple poles at each non-zero integer, (c) has simple poles at all of the integers.

How do we conclude that an infinite sum of meromorphic functions is meromorphic? Do I need to show that they converge? Why is this enough?

Also, if the numerator of (b) were $\sin(\pi z)$ would this change the order/existence of the poles?

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"Do I need to show that they converge?" - yes. –  J. M. Jul 21 '11 at 7:41
    
"Also, if the numerator of (b) were $\sin(n\pi z)$ would this change the order/existence of the poles?" - if $z$ were rational, then there exists an $n$ where the expression is 0, right? –  J. M. Jul 21 '11 at 7:44
    
Why is showing convergence enough? For (b), what are the implications of that. For each $z$, a bunch of the terms would be $0$, but not all of them. –  RHP Jul 21 '11 at 7:55
    
I think what I meant was: what if the numerator were $\sin(\pi z)$. (How) would this change what is going on at the integers? –  RHP Jul 21 '11 at 7:56
5  
Use the fact that infinite sum of holomorphic functions (on a domain) that converge uniformly on every compact subset is holomorphic. In this case, consider a disk of radius R. Apply what I said previously to a suitable tail of the series. You are left with a finite sum of meromorphic functions, which is of course meromorphic. –  Soarer Jul 21 '11 at 8:05
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up vote 3 down vote accepted

How do we conclude that an infinite sum of meromorphic functions is meromorphic? Do I need to show that they converge? Why is this enough?

The answer is essentially contained in Montel's theorem. Let's say we have a sum of meromorphic functions: $$ \sum_{n = 0}^{\infty} f_n $$ First question: Can we find a domain $D \subseteq \mathbb{C}$ so that none of the poles of any $f_n$ is contained in $D$? If yes, then the next question is if the series $$ F_N := \sum_{n = 0}^{N} f_n $$ is locally bounded on $D$ (this is the part where you need to prove the convergence of the series in the sense that you need to show that there is a finite upper bound for all $z \in D$). If it is, we can apply Montel's theorem and conclude that $F_N$ is normal, that is it has a subsequence that converges compactly to a holomorphic function $F$ on $D$. Since the sequence $(F_N)$ is a Cauchy sequence, we see that actually $(F_N)$ itself converges to $F$ on $D$.

In order to show that $F$ is meromorphic on $\mathbb{C}$, we need to show that $F$ does not have an essential singularity and that the set of poles is a discrete subset of $\mathbb{C}$.

This will work out if you can show that

  • $F$ is finite in every z that is not a pole of any $f_n$,

  • the set of all poles of all $f_n$ is discrete and

  • for every z that is a pole of some $f_n$, z is a pole of only finite many $f_n$.

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