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Writing down some easy rational functions to check this, I don't see why this must be the case.

Although if the function had 3 simple zeros and 2 simple poles its rational form would be in the form of a $\frac{cubic}{quadratic}$, and this function doesn't stay bounded near infinity.

However, flipping the above, i.e., let's say $f$ has 2 simple zeros and 3 simple poles, and is required to stay bounded near infinity.

Doesn't this function exist? It would be a $\frac{quadratic}{cubic}$, which stays bounded near infinity.

A solution that I am reading claims that such a function cannot exist, unless the number of zeros and poles are the same.

Any ideas are welcome.

Thanks,

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    $\begingroup$ reading where?? $\endgroup$
    – Will Jagy
    Dec 20, 2015 at 21:44
  • $\begingroup$ Solutions from previous students in our math dept. I have found plenty of mistakes in reading their solutions, when I go to compare my work with theirs, but also have read some good solutions, too. So it's a mixed bag. I wouldn't be surprised if this solution is just plain wrong. $\endgroup$
    – User001
    Dec 20, 2015 at 21:52
  • $\begingroup$ And that the real claim is that, if the function is required to have three simple zeros, 2 simple poles, and is to be bounded near infinity, then such a function cannot exist (because it cannot stay bounded near infinity) The mentioning of the requirement for the number of poles and zeros to be the same is probably not correct. What do you think? Thanks, @WillJagy $\endgroup$
    – User001
    Dec 20, 2015 at 21:52
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    $\begingroup$ Mostly I think that a nonzero limit at infinity requires the numerator and denominator to be the same degree, while just bounded at infinity does not. $\endgroup$
    – Will Jagy
    Dec 20, 2015 at 21:54
  • $\begingroup$ Yes, good point, and I agree with you. Hmm...I'll just ignore what I had read and move on to the next problem then. Thanks for your help, @WillJagy, $\endgroup$
    – User001
    Dec 20, 2015 at 21:59

2 Answers 2

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The idea is that the zero or pole at infinity cancels out the finite zeroes and poles. For example, in your $quadratic/cubic$ example, there are two finite zeroes, three finite poles, and a simple zero at infinity. The proof is as follows: If $f(z)=\dfrac{z^2+az+b}{z^3+cz^2+dz+e}$, then $f(1/z)=\dfrac{1+az+bz^2}{1+cz+dz^2+ez^3}z$ which clearly has a simple zero at $z=0$. This proof generalizes easily to any rational function, and thus any meromorphic function on the Riemann sphere.

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  • $\begingroup$ Wow zeros at infinity. Nice. And the order difference of the polynomials is the multiplicity of the zero at infinity? $\endgroup$
    – mathreadler
    Dec 20, 2015 at 22:24
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Noah Olander
    Dec 20, 2015 at 22:46
  • $\begingroup$ Maybe I'm just stupid but I don't understand this (written after a few minutes thinking about your text and after rush upvoting). What does it prove? $\endgroup$
    – Peter Franek
    Dec 20, 2015 at 22:50
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    $\begingroup$ Oh, all meromorphic functions because every meromorphic function on the Riemann sphere is rational. It's a famous theorem, you can find it in Stein and Shakarchi for instance, and surely elsewhere too. $\endgroup$
    – Noah Olander
    Dec 20, 2015 at 23:06
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    $\begingroup$ Yeah if the numerator is degree three and the denominator is degree two, you get a simple pole at infinity, so the function isn't bounded at infinity. To summarize, the proof consists of (a) writing out the definition of f (f must be rational by an above comment), (b) plugging in $1/z$ for $z$, (c) multiplying the equation by 1 cleverly, in such a way that you you see $f(1/z)=\frac{p(z)}{q(z)}z^k$ where $p,q$ are polynomials and $p(0) \neq 0 \neq q(0)$. Now you can read off the singularity of $f(1/z)$ at zero, which by definition is the singularity of $f$ at infinity. $\endgroup$
    – Noah Olander
    Dec 21, 2015 at 0:36
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The number of zeroes is equal to the number of poles (counted with multiplicities) for any meromorphic function on a (compact, connected) Riemann surface.

Taken from these notes on Rieman-Roch theorem.

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    $\begingroup$ Hi @PeterFranek, I am not so sure about your claim. Specifically, a meromorphic function with polynomial growth near infinity is not required to have the same number of poles and zeros - in fact, I think it must have more zeros than poles (to satisfy the polynomial growth condition.) $\endgroup$
    – User001
    Dec 20, 2015 at 22:20
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    $\begingroup$ @User001 Right, it only holds for compact surfaces (similarly as in Rieman-Roch theorem. In the text I refer to, compactness is used in the Proof of 3.4). But the Riemannian sphere is compact. If your meromorphic function on $\Bbb C$ has a limit in $\infty$ then it can be extended to a meromorphic function on the Riemannian sphere. $\endgroup$
    – Peter Franek
    Dec 20, 2015 at 22:30
  • $\begingroup$ Ok, got it -- thanks so much @PeterFranek :-) $\endgroup$
    – User001
    Dec 21, 2015 at 1:40

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