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Some Complex Analysis help

  1. Mar 16, 2006 #1
    Here's my question:

    Let f and g be analytic inside and on the smple loop [tex]\Gamma[/tex]. Prove that if f(z)=g(z) for all z on [tex]\Gamma[/tex], then f(z)=g(z) for all z inside [tex]\Gamma[/tex].

    Don't really know where to start on this one. This comes from the section 'Cauchy's Integral Formula'.
     
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  3. Mar 16, 2006 #2

    cepheid

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    What does Cauchy's Integral Formula tell you about the value of a function at point z0 **inside** Gamma in terms of the integral of that function around Gamma?
     
  4. Mar 16, 2006 #3
    well it states that for a point z0 inside gamma,

    [tex]f(z_0)=\frac{1}{2\pi\i}{\int_\Gamma}\frac{f(z)}{z-z_0}dz[/tex]

    But where do I go from there. I think i'm missing something big and obvious :x

    k here is my guess:

    since [tex]f(z_0)=g(z_0)[/tex] for all [tex]z_0[/tex] on [tex]\Gamma[/tex], then we know

    [tex]\frac{1}{2\pi\i}{\int_\Gamma}\frac{f(z)}{z-z_0}dz=\frac{1}{2\pi\i}{\int_\Gamma}\frac{g(z)}{z-z_0}dz[/tex], and therefore f(z)=g(z) for all z in [tex]\Gamma[/tex] since this formula is ambiguous whether z_0 is on [tex]\Gamma[/tex] or inside of it.

    Problem is this seems like i'm missing something. For this case, even if I write that shouldn't it still only apply to the loop itself and not what is inside. Is there some subtle thing I am missing that would allow me to actually say what I just said?
     
    Last edited: Mar 16, 2006
  5. Mar 16, 2006 #4

    HallsofIvy

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    That's a perfectly valid proof. However, it's not true that "this formula is ambiguous whether z_0 is on or inside of it". It requires that z0 be inside the loop so that [itex]\frac{f(z)}{z-z_0}[/itex] be defined for all x on the loop.
     
  6. Mar 17, 2006 #5

    cepheid

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    How can you start the proof with this statement? That is what you are trying to prove!

    Ummmm......NO. As Halls pointed out, z0 is a point INSIDE gamma, not on it.

    I think this is the statement you start the proof with. It follows immediately from the fact that f(z) = g(z) on Gamma. From there the proof is as straightforward as applying the formula.
     
  7. Mar 17, 2006 #6
    Sorry, I don't think I was thinking too clearly when I wrote that. Thanks for the advice.
     
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