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

In summary, to prove that if f(z) is analytic over a simply connected domain containing a simple closed curve C and Z_{0} is a point inside C, then f'(z_{0}) = \frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz, the Cauchy Integral Formula must be used. However, the proof requires knowledge of the first version of the Cauchy Integral Formula, and it is important to be careful when permuting the limit and integral signs. If this proof is needed for an exam, it is recommended to review the key theorems and have a general outline of the proof in order to be
  • #1
gtfitzpatrick
379
0

Homework Statement



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]

Homework Equations





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?
 
Physics news on Phys.org
  • #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.
 
  • #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?
 
  • #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..."
 
  • #5
its for an exam, could possibly come up. But the proof is very long to learn off! :(
 
  • #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!
 

1. What is the Cauchy integral formula?

The Cauchy integral formula is a fundamental result in complex analysis that allows us to calculate the value of a complex function at any point inside a simple closed curve, given its values on the boundary of the curve.

2. How is the Cauchy integral formula derived?

The Cauchy integral formula is derived from Cauchy's integral theorem, which states that if a function is analytic within a closed curve, then the integral of the function around that curve is equal to 0. By using this theorem and some complex analysis techniques, we can derive the Cauchy integral formula.

3. What is the significance of the Cauchy integral formula?

The Cauchy integral formula is significant because it allows us to extend the concept of differentiation to complex functions and evaluate integrals involving these functions. It is also a powerful tool for solving various problems in physics and engineering.

4. Can the Cauchy integral formula be applied to any curve?

No, the Cauchy integral formula can only be applied to simple closed curves, which are curves that do not intersect themselves and enclose a single, connected region.

5. Is the Cauchy integral formula enough to prove the existence of a complex function?

No, the Cauchy integral formula is not enough to prove the existence of a complex function. It is only one of the many tools used in complex analysis to study and analyze complex functions. Other techniques, such as the Cauchy-Riemann equations and the uniqueness theorem, are also needed for a complete proof of the existence of a complex function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
847
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Topology and Analysis
Replies
2
Views
616
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
943
  • Topology and Analysis
Replies
14
Views
441
  • Calculus and Beyond Homework Help
Replies
3
Views
534
  • Calculus and Beyond Homework Help
Replies
19
Views
965
  • Calculus and Beyond Homework Help
Replies
3
Views
846
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top