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        general >> complex analysis >> meromorphic function
        
(Message started by: Moon on Apr 10^th, 2004, 6:49pm)

Title: meromorphic function
Post by Moon on Apr 10^th, 2004, 6:49pm

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,...,P_k} and suppose that f has a pole at each of the points P₁,P₂,...,P_k. 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.

Title: Re: meromorphic function
Post by Icarus on Apr 11^th, 2004, 6:26pm

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]?

Title: Re: meromorphic function
Post by Moon on Apr 12^th, 2004, 3:49pm

f/R has poles at P₁, P₂,...,P_k, p₁, p₂,...,p_n, where p_n are zeros of p(z). Also we can state that f has zeros at p₁, p₂,...,p_n, 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.

Title: Re: meromorphic function
Post by Icarus on Apr 12^th, 2004, 4:56pm

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.

Title: Re: meromorphic function
Post by Moon on Apr 12^th, 2004, 6:38pm

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.

Title: Re: meromorphic function
Post by Icarus on Apr 13^th, 2004, 6:09pm

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.

Title: Re: meromorphic function
Post by Moon on Apr 13^th, 2004, 7:23pm

Bounded entire function is a constant.

Thank you very mauch for your help.