Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm doing some exercises in complex analysis, and I've reached one I simply can't figure out on my own, which is why I'm hoping for some help.

The exercise:

We assume that $h:\Bbb C\to \Bbb C \cup \{\infty\}$ is meromorphic with finitely many poles $z_1,...z_n$ and assume that there exist $k\gt0$, $N\in\Bbb N$ and $R\gt0$ such that $|h(z)|\le k|z|^N$ for $|z|\gt R$. Prove that h is a rational function.

What I've been thinking so far:

The definition of a rational function is, that you have to be able to write it on the form $f(z)=p(z)/q(z)$, where $p,q\in\Bbb C[z], q\neq 0$. So I guess I have to show that my function $h$ can be written this way too? I have, however not yet been able to figure out a way to do this which makes sense.

Earlier we've done an exercise, in which we've proven that if we let f be an entire function and we assume that $|f(z)|\le A+B|z|^n$ for $z\in\Bbb C$, where $A,B\ge0$ and $n\in\Bbb N$, then f is a polynomial of degree $\le n$. I've been thinking this might be useful, although a meromorphic function isn't entire.

Thank you for your time.

share|improve this question
5  
Hint: You're right, but you can find a polynomial to multiply by that will "cancel" all the poles and leave you with an entire function. –  Ted Shifrin May 24 '13 at 20:04
add comment

1 Answer

up vote 2 down vote accepted

Since $h$ has a finite number of poles, $h(z)\cdot (z-z_1)\cdots (z-z_n)$ is an entire function. Let $w_1,\ldots,w_n$ (this could be an empty list) be the zeros of $h(z)\cdot (z-z_1)\cdots (z-z_n)$. There must be a non-zero entire function $f$ such that $$ h(z)\cdot (z-z_1)\cdots (z-z_n)=(w-w_1)\cdots (w-w_n)\cdot f(z) .$$

We show that $f$ must be a constant and the problem will be solved.

By assumption $$|(w-w_1)\cdots (w-w_n)\cdot f(z)|\le k|z|^N\cdot |(z-z_1)\cdots (z-z_n)|\le j|z|^M$$ for some $j$ and $M$.

Thus $(w-w_1)\cdots (w-w_n)\cdot f(z)$ must be a polynomial and consequently $f$ is a non-zero polynomial (i.e. a constant).

share|improve this answer
    
This is "morally" correct, but not correct as it stands. The first sentence needs to be repaired, with corresponding changes throughout. –  Ted Shifrin May 25 '13 at 12:50
    
Not sure what you mean. Do mean I am using "pole" too slangishly? There are functions with an infinite number of poles. Also, there are functions with uncountable singularities. –  Bobby Ocean May 26 '13 at 5:05
1  
Multiplying by linear terms only kills single poles. If you have $1/z^2$, for example, you need to multiply by $z^2$, not just $z$. If you multiply by $\prod(z-z_j)^{n_j}$ instead of $\prod(z-z_j)$, where $n_j$ is the order of the pole, you do get an analytic function, and you can proceed as you indicated. –  Potato May 26 '13 at 5:22
    
Yes of course; Sorry I assumed the list $z_1,\ldots,z_n$ included repeating the poles, which is not a standard assumption. I should have said, let $z_1,\ldots,z_n$ be the poles including multiplicity. –  Bobby Ocean May 27 '13 at 17:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.