Do zeros of meromorphic functions have a limit point?

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Homework Help Overview

The discussion revolves around the properties of meromorphic functions on the complex plane, specifically focusing on the nature of their zeros and whether they can have a limit point. Participants are exploring the implications of meromorphic functions being holomorphic except at isolated poles and the consequences of having zeros in relation to the identity theorem and the Pigeon Hole Principle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the potential application of the identity theorem and the implications of having a set of zeros with a limit point. There are attempts to connect the concept of dividing the complex plane into squares to demonstrate the distribution of zeros and the use of the Pigeon Hole Principle to argue about the countability of zeros.

Discussion Status

The discussion is active with various approaches being considered. Some participants are questioning how to apply the identity theorem effectively, while others are exploring the implications of dividing the complex plane into squares. There is a recognition of the need to show that zeros must have a limit point, indicating a productive direction in the exploration of the problem.

Contextual Notes

Participants are working under the constraints of the properties of meromorphic functions and the definitions of zeros and poles. The discussion includes assumptions about the distribution of zeros and the implications of having uncountably many zeros in relation to the boundedness of sets in the complex plane.

Dustinsfl
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Let [itex]f[/itex] be a nonzero meromorphic function on the complex plane. Prove that [itex]f[/itex] has at most a countable number of zeros.

Since [itex]f[/itex] is meromorphic on [itex]\mathbb{C}[/itex], [itex]f[/itex] is holomorphic on [itex]\mathbb{C}[/itex] except for some isolated singularities which are poles. Aslo, [itex]f[/itex] being meromorphic we can write [itex]f[/itex] as [itex]g(z)/h(z)[/itex] both holomorphic with [itex]h\neq 0[/itex].

Now does multiplying through help lead to the conclusion?

So we would have [itex]fh = g[/itex]. If so, I am not sure with what to do next.
 
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What about using the identity theorem?? If two holomorphic functions agree on a set with a limit point, then they are equal.
 
micromass said:
What about using the identity theorem?? If two holomorphic functions agree on a set with a limit point, then they are equal.

I understand what you are saying but not sure on how to apply it. Should I start over?
 
Dustinsfl said:
I understand what you are saying but not sure on how to apply it. Should I start over?

Divide the complex plane up into squares. If there were an uncountable number of zeros then one of the squares must contain an infinite number of zeros. Can you prove that?
 
Dick said:
Divide the complex plane up into squares. If there were an uncountable number of zeros then one of the squares must contain an infinite number of zeros. Can you prove that?

I could probably use the Pigeon Hole Principle to show that but how would that help?
 
Dustinsfl said:
I could probably use the Pigeon Hole Principle to show that but how would that help?

Then use micromass's suggestion. Can you show the zeros must have a limit point?
 
Dick said:
Then use micromass's suggestion. Can you show the zeros must have a limit point?

Suppose [itex]f[/itex] has uncountably many zeros and divide the complex plane into squares of with edges lengths of 1 centered at [itex]a + bi[/itex] where [itex]a,b\in\mathbb{Z}[/itex]. By the Pigeon Hole Principle, one of these squares contains uncountably many zeros. Now that we have infinitely many zeros in a bounded set, we have a limit point. By Liouville's Theorem, [itex]f[/itex] must be constant. Therefore, [itex]f[/itex] has at most countably many zeros.
 
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