Proving this meromorphic function has a primitive in punctured plane

In summary, the conversation discusses how to prove that the function f(z) - \frac{b_{-1}}{z-a} has a primitive in a punctured neighborhood of a, where f(z) is analytic everywhere except at a singularity. The solution involves using the Laurent series of f(z) and showing that it is uniformly convergent on compact sets, which allows for the integral to be pushed through the infinite series and integrated term by term. This proves that the integral of the function along any closed curve in the punctured plane is zero, and therefore f(z) has a primitive on the entire punctured plane.
  • #1
Poopsilon
294
1

Homework Statement



Let [itex]f(z)[/itex] be a complex function analytic everywhere except at [itex]a[/itex] where it has a singularity. Prove that the function [itex]f(z) - \frac{b_{-1}}{z-a}[/itex] has a primitive in a punctured neighborhood of [itex]a[/itex]. Where [itex]b_{-1}[/itex] is the coeffecient of the n=-1 term in the Laurent expansion of [itex]f(z)[/itex].

Homework Equations



Well I know Laurent series are compactly convergent. I also know path independence of an arbitrary line integral on a punctured neighborhood of [itex]a[/itex] would imply a primitive, so would having the integral over all closed curves equaling zero.

The Attempt at a Solution



What I'd really like to do is just expand f(z) into its Laurent series, subtract off the [itex]\frac{b_{-1}}{z-a}[/itex] term, and then push the integral operator through the infinite series and integrate term by term. But I think I would need uniform convergence for that, and not just uniform convergence on compact subsets.

I feel like if I had a deeper understanding of uniform convergence I might see why only having compact convergence would be sufficient, but I don't. Beyond that I'm stumped. Help would be much appreciated, thanks.
 
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  • #2
Since f(z) is analytic everywhere except at α, doesn't that mean the Laurent series is uniformly convergent in the entire punctured complex plane? Correct me if I'm wrong.
 
  • #3
According to wikipedia it is uniformly convergent on compact sets and thus on only and all sets which are closed and bounded. So I think you would have to put a little open ball around the singularity and then f(z) will be uniformly convergent outside of that ball.
 
  • #4
but any closed curve in the punctured plane should be contained in the plane less a small enough open ball around a, is that right?
 
  • #5
Yes you're right, so the question is, is the quantifier(existential and all) situation such that given an arbitrary closed curve in the punctured plane we are allowed to choose an open ball small enough that it cuts out the singularity while leaving the curve untouched, and then at this point we can push the integral along this closed curve through the infinite series and integrate term by term which will give us zero and therefore say something to the effect of:

The integral of this function along any closed curve in the punctured plane is zero and thus f(z) has a primitive on the entire punctured plane.

It seems right but at the same time I feel like I might be missing something. Because you can always keep choosing curves nearer and nearer to the singularity and would keep having to expand your set on which the series converges uniformly and I just can't tell which you are allowed to choose and which must be given, hope this makes sense.
 
  • #6
Obviously [itex]F(z)=\int^z f(z)dz[/itex] is well defined in the entire punctured plane, and F'(z)=f(z), q.e.d. Looks OK to me :)
 
  • #7
Ya that seems right, since what is given to you is the arbitrary curve and then all you're really concerned with is whether the integral along it is zero and since you can always find that compact set, you're good to go. Thanks for your help.
 

Related to Proving this meromorphic function has a primitive in punctured plane

What does it mean for a function to be meromorphic?

A meromorphic function is a complex-valued function that is analytic everywhere except for isolated points where it has poles. This means that it can be expressed as a ratio of two analytic functions.

What is a primitive of a meromorphic function?

A primitive of a meromorphic function is an analytic function whose derivative is equal to the given meromorphic function. In other words, it is the inverse operation of taking derivatives.

Why is it important to prove that a meromorphic function has a primitive in the punctured plane?

Proving that a meromorphic function has a primitive in the punctured plane is important because it allows us to integrate the function along any path in the punctured plane. This is useful in many areas of mathematics and physics, such as in complex analysis and in calculating residues.

What are the steps involved in proving that a meromorphic function has a primitive in the punctured plane?

To prove that a meromorphic function has a primitive in the punctured plane, we first need to show that the function is analytic everywhere except for isolated points. Then, we can construct a primitive by finding a function whose derivative is equal to the given meromorphic function. This can be done by using the Cauchy integral formula or by finding a suitable antiderivative. Finally, we need to show that this primitive is unique up to a constant.

What are some common techniques used in proving that a meromorphic function has a primitive in the punctured plane?

Some common techniques used in proving that a meromorphic function has a primitive in the punctured plane include using the Cauchy integral formula, finding a suitable antiderivative, and using properties of analytic functions such as the Cauchy-Riemann equations. Other techniques may include using residue calculus or the theory of complex integration.

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