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Proving this meromorphic function has a primitive in punctured plane

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    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].


    2. Relevant 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.



    3. 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.
     
  2. jcsd
  3. Mar 17, 2012 #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.
     
  4. Mar 17, 2012 #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.
     
  5. Mar 17, 2012 #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?
     
  6. Mar 17, 2012 #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.
     
  7. Mar 17, 2012 #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 :)
     
  8. Mar 17, 2012 #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.
     
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