# Homework Help: A problem related with a meromorphic function

1. Mar 22, 2012

### gotjrgkr

1. The problem statement, all variables and given/known data
Hello!
I have a question about a meromorphic function introduced in a field of complex analysis.
I refer to palk's book "an introduction to complex function theory".
As you can check in that book, the author introduces a definition of a meromorphic function in p. 318. I quote the definition here; " We characterize a function f as meromorphic in an open set U provided f has at no point of U worse than a pole.". you can also find the expression " a function f has no worse than a pole at a point z$_{0}$" meaning that there exists a radius r > 0 such that the given function f is either analytic in the full open disk Δ(z$_{o}$,r) or else analytic in the punctured disk Δ$^{\ast}$(z$_{0}$,r) with a pole or removable singularity at its center. You can check this in p. 317 in the book.

Now, if you look at a paragraph right above "2.5 Essential Singularities" in p. 319, it is said that if functions f and g are both meromorphic in an open set U, then you can get directly from the th.2.5 in p.318 f/g are likewise meromorphic in this open set unless g is identically zero in some component of U.

I think I misunderstand something here, so that I can't see the result.

2. Relevant equations

3. The attempt at a solution
I've tried to apply th. 2.5 in p. 318 to prove the remark witten in p. 319 which I've written right above. I think there would be a problem when g has a removable singularity at z$_{0}$ with the limit limit$_{z\rightarrowz_{0}}$g(z)=0 where z$_{0}$ is a point of a component of U. What I mean is I can't safely say that g is defined in a punctrued disk around z$_{0}$. I think I need to gaurantee (from the assumption g is not identically zero in any component of U) that for any point in a component of the open set U which is a removable singular point with limit$_{z\rightarrowz_{0}}$g(z)=0, there's a punctured disk around that point where f is not identically zero, so that I can apply corollary 1.3 in p. 302 to say safely that f/g is well defined in a punctured disk centered at z$_{0}$.

But, I can't prove it. Furthermore, I expect there's a example such that under the assumption g is not identically zero in any component of U, f/g can't be well defined.
; Let U be an open disk centered at 0, g be a function whose domain set is the disk and its value is one at zero and zero otherwise. (f is any analytic function in the disk)

2. Mar 28, 2012

### morphism

Having $\lim_{z\to z_0} g(z) = 0$ is no problem. If g is identically zero in a punctured neighborhood of z_0, then it will be identically zero in the connected component of U containing z_0 (by the identity theorem), which cannot happen. Thus g is nonzero in some punctured disk about z_0. So you can "remove" this singularity and redefine g so that g(z_0)=0. Hence we can conclude that f/g has at worst a pole at z_0.

Does this help?

3. Apr 7, 2012

### gotjrgkr

Oh, I'm sorry. I've just seen it.
I find "the identity theorem" by searching it through google.
It is said that
"In complex analysis, a branch of mathematics, the identity theorem for holomorphic functions states: given functions f and g holomorphic on a connected open set D, if f = g on some neighborhood of z that is in D, then f = g on D. Thus a holomorphic function is completely determined by its values on a (possibly quite small) neighborhood in D.".
I know what you mean when you use the theorem to prove it in the case of limit$_{z\rightarrowz_{0}}$g(z)=0. But, I'm still confused. As you can see from above theorem which you suggested, in order to apply the theorem you need to confirm that g is analytic in the component containg z$_{0}$ of U. But, I can think of an example such that g is not analytic in the component containing z$_{0}$, satisfies limit$_{z\rightarrowz_{0}}$g(z)=0, is not identically zero in the component, and is zero in a punctured disk centered at z$_{0}$.
That is, Let g be a function whose domain set is an open disk centered at zero and whose values are 1 at zero, 0 if z is not zero and belongs to the disk. In this case, g is not analytic in the whole disk. You can also check that the only component of the disk is the disk itself.
You can't apply "the identity theorem" to this example because g is not analytic in the whole disk. I, however, find that the function is meromorphic in the domain set. Its only singularity is the removable singularity at 0. If I take any analytic function in the domain set, then f/g can't be meromorphic in the domain... ;;
So, a condition that g is not identically zero in any component of U is not sufficient to show that f/g is meromorphic in U.
Is there anything that I misunderstands in the above statement??
Could you tell me which part??

4. Apr 8, 2012

### Office_Shredder

Staff Emeritus
You seem to be confused about the definition of meromorphic! The function is required to be analytic away from the poles

Also your function has no singularities, it's just discontinuous

5. Apr 8, 2012

### gotjrgkr

I refer to a book " an introduction to complex function theory" by palka.
In p. 318 of the book, a meromorphic function is defined as follows;
Let U be an open set in the complex plane. We characterize a function f as meromorphic in U provided f has at no point of U worse than a pole.
" no worse than a pole" means that there's a radius r>0 such that the given function f is either analytic in the full open disk Δ=Δ(z$_{0}$,r) or else analytic in the puntured disk Δ$^{*}$(z$_{0}$,r) with a pole or removable singularity at its center.

I think the book you use is different from mine.
I thought that since the function g I suggested above is analytic in a disk centered at each point except for zero, g is also analytic in a punctured disk centered at zero, and all negative terms of its laurent expansion centered at zero are 0 , I thought g has a removable singularity at 0 and g is meromorphic in the disk. Am I wrong?? If so, could you tell me which part is wrong??

6. Apr 8, 2012

### Office_Shredder

Staff Emeritus
A removable singularity is something like f(z)=z/z, which isn't defined at 0 because thre denominator is zero but f(z) had a limit as z goes to 0 so you can remove the singularity by picking a value for f at 0. Your function has a discontinuity which is not the same thing