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Cauchy integral formula(is this enough of a proof?)

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if f(z) is analytic over a simply connected domain containing a simple closed curve C abd Z[itex]_{0}[/itex] is a point inside C then f'(z[itex]_{0}[/itex]) = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz[/itex]

    2. Relevant equations



    3. The attempt at a solution
    from the definition of a derivative we have f'(z[itex]_{0}[/itex]) =[itex]\stackrel{lim}{h\rightarrow0} \frac{f(z_0 +h) - f(z_0)}{h}[/itex]
    the cauchy integral formula states f(z[itex]_{0}[/itex]) = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)} dz[/itex] and so f(z[itex]_{0}+h[/itex]) = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0 +h)} dz[/itex]

    then subing the last 2 into the first one i get f'(z[itex]_{0}[/itex]) = [itex] \stackrel{lim}{h\rightarrow0}[/itex] [itex] \frac{1}{2i\pi}[/itex] [itex] \oint_{c} [/itex] [itex] \frac{1}{h} [/itex] ([itex] \frac{f(z)}{(z- z_0 - h)}[/itex] [itex] - \frac{f(z)}{(z-z_0)} )dz[/itex]

    =f'(z[itex]_{0}[/itex]) = [itex] \stackrel{lim}{h\rightarrow0}[/itex] [itex] \frac{1}{2i\pi}[/itex] [itex] \oint_{c} [/itex] ([itex] \frac{f(z)}{(z- z_0 - h)(z-z_0)}[/itex] )dz
    on[/b]

    = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz[/itex]

    is this enough of a proof or would i also have to prove cauchys integral formula?
     
  2. jcsd
  3. Aug 21, 2012 #2
    The proof looks fine, except for two things:

    1. It assumes you already know the first version of the Cauchy Integral Formula (eg. [itex]f(z_0)=\frac{1}{2\pi i}\int_C \frac{f(z)}{z-z_0}dz[/itex]). I might be misunderstanding your question at the bottom, but if you haven't already proven this result, then you must do so before invoking it in your proof.

    2. One line before your conclusion, you permute the limit with the integral. Are you sure you can do that? By uniform convergence, if
    [tex]g_n(z)=\frac{f(z)}{(z-z_0-h_n)(z-z_0)}[/tex]
    converges uniformly to
    [tex]
    \frac{f(z)}{(z-z_0)^2}[/tex]
    for every sequence [itex] h_n\rightarrow 0[/itex], then you can permute the limit and the integral sign. Do you have this result? If you aren't looking for a rigorous proof, then what you have is enough. However, in general, you cannot permute the integral and limit like that.
     
  4. Aug 21, 2012 #3
    hi christoff,
    Thanks for reply. The reson why i didnt put in the Proof of the Cauchy integral formula as it is very long and hard to remember. Do you know anywhere i can get a hardy version that i could throw in?
     
  5. Aug 21, 2012 #4
    That really depends on what this is for. If this is for homework, then any complex analysis textbook will cover it, and you should be able to find a proof and figure it out on your own. In that case however, you've likely covered the (first) integral formula in class, so you are probably free to say "by the (first) Cauchy Integral Formula..."
     
  6. Aug 21, 2012 #5
    its for an exam, could possibly come up. But the proof is very long to learn off! :(
     
  7. Aug 21, 2012 #6
    Ah, that's unfortunate. I remember when I was taking my second course in real analysis, our professor offered us a recommended study guide for the final exam. It included the suggestion: "Know how to prove all of the key theorems". Very helpful.

    In that case, you will have to weigh the risks and decide if you want to learn the proof, or skip it and hope that it doesn't come up in an exam. In that situation, I like to review the proof enough to come up with an outline - not necessarily every little step, but enough of a "skeleton" so that I can hopefully figure it out in the moment. It works for me...

    Good luck!
     
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