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Topic: meromorphic function (Read 5985 times) |
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Moon
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please help me with this problem:
Let R(z) be a rational function: R(z)=p(z)/q(z) where p and q are holomorphic polynomials. Let f be holomorphic on C\{P1,P2,...,Pk} and suppose that f has a pole at each of the points P1,P2,...,Pk. Finally assume that |f(z)|<=|R(z)| for all z at which f(z) and R(z) are defined. Prove that f is a constant multiple of R. In particular, f ia rational. (Hint: Think about f(z)/R(z).)
Thank you in advance.
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Icarus
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Re: meromorphic function
« Reply #1 on: Apr 11th, 2004, 6:26pm » |
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First of all, I have to point out that "holomorphic polynomial" is redundant. All polynomials are holomorphic.
What kind of singularities can f/R have? Once you have answered that (it should be easy to figure out from the info you've been given), what does the inequality |f/R| [le] 1 tell you about these singularities?
Lastly, what are the possibilities for bounded functions holomorphic on all of [bbc]?
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Moon
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f/R has poles at P1, P2,...,Pk, p1, p2,...,pn, where pn are zeros of p(z). Also we can state that f has zeros at p1, p2,...,pn, but I can't still answer the question how inequality |f/R|<=1 may state anything about singularities. My intuition tells me that from |f/R|<=1 follows that f/R is entire and since it's bounded then by Liouville Theorem follows that it's a constant.
Please help me to answer this question.
Thank you.
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Icarus
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Re: meromorphic function
« Reply #3 on: Apr 12th, 2004, 4:56pm » |
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What does "pole" mean? What happens to the values of the meromorphic function f/R as z approaches a pole? How can you reconcile this with the inequality |f/R| [le] 1?
The inequality puts a very strong condition on how f and R are related.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Moon
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Am I correct that as z approaches a pole f/R goes to infinity. So there exists such neighbourhood at any pole such that any point z in this neighbourhood gives us that |f/R(z)|>1. This means that no poles in the domain of f/R, hence f/R is entire.
Correct me if I'm wrong please.
Thank you.
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Icarus
wu::riddles Moderator
Uberpuzzler
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Re: meromorphic function
« Reply #5 on: Apr 13th, 2004, 6:09pm » |
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Exactly! f/R has no essential singularities because neither f nor R have them. It has no poles because of the |f/R| [le] 1 condition. So all of its singularities must be removable. Hence my final hint:
What are the possibilities for bounded functions holomorphic on all of [bbc]?
This is a result you should have seen by now, and can look up in your textbook.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Moon
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Bounded entire function is a constant.
Thank you very mauch for your help.
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