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Please help w/ entire functions

  1. Feb 27, 2008 #1
    Context: this is complex anaylsis II and I can use:

    Cauchy's integral theorem
    Liouville's Theorem
    Taylor's Theorem
    Morera's Theorem

    ---

    Let's say you have a function like g(z) = f(z)/z

    And you know that f(z) is entire. But, then you find out that

    [tex]\displaystyle\oint_{c} g(w) dw =0[/tex]

    c is a closed curve

    that implies that g(z) is analytic in the region where this occurs, the region includes zero. But, how could g(z) be analytic? there is some kind of singularity at z=0. Could it just be removable? I'm not happy with this.
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2
    I'm pretty sure that what you have is

    [tex]\displaystyle\oint_{c} g(w) dw = \displaystyle\oint_{c} \frac{f(w)}{w} dw = 2 \pi i * n(c, 0)*f(0)[/tex]

    So for that to be zero for all closed curves, f(0) = 0. Dunno if that helps any. And, as always, I could be wrong.
     
  4. Feb 27, 2008 #3
    thanks!

    That happens to be how I got to this point. I want to say g(z) is entire, but I'm uset by the singularity at zero. If g(x) was something like...

    3z/z can one say that it is "entire" even though it has no defined value at zero?
     
  5. Feb 27, 2008 #4
    Wow, I'm really slow today. Okay, g(z) is not entire because, as you said, it's not analytic at 0. Why does your argument fail? Because the integral [itex]\displaystyle\oint_{c} g(w) dw[/itex] is not defined if c passes through 0, so it's not 0 for every closed curve in the complex plane. However, since z*g(z) goes to 0 as z goes to 0 (because f(0) = 0 and f is continuous), the singularity at 0 is removable. So you could extend g to an analytic function on the whole plane, and that's probably good enough.
     
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