Questions about complex analysis (Cauchy's integral formula and residue theorem)

In summary: Instead, we just needed to show that the interior of the gamma function is open. In other words, we could write f(a+h) as a contour integral around g, and then take the difference quotient. Is this right?
  • #1
gangsta316
30
0
http://www2.imperial.ac.uk/~bin06/M2...nation2008.pdf [Broken]

Solutions are here.

http://www2.imperial.ac.uk/~bin06/M2...insoln2008.pdf [Broken]

My first question is about 3(ii), the proof of Cauchy's integral formula for the first derivative.

The proof here uses the deformation lemma
(from second page here:
http://www2.imperial.ac.uk/~bin06/M2...pm3l18(11).pdf [Broken])
and proves the theorem for an approximating contour. I made up the proof myself using the ideas from what we were taught (so I remembered the gist of the proof, not all of it) and I think that I made one without the use of this lemma. Why is it needed? Can we not just say that, since the interior of g (g for gamma) is open, a+h is inside g for |h| small enough. Then we can write f(a+h) as a contour integral around g and then take the difference quotient and let h->0. Is this ok or do we need to be working in a circular contour for the proof to work? I always thought something was not quite right about how he just let's h->0 inside the integrand like that (i.e. without evaluating the integral) -- is that what requires the contour to be a circle?


4(ii).
Would it be correct to substitute x = tany and change the limits to -pi/2, +pi/2? In general how do we know that these will be the limits for this integral? tan also blows up at (pi/2 + 2*pi) so why did we choose pi/2 for the upper limit?

Thanks for any help.
 
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  • #2
Those links are not working.
Like much of calculus this is a problem with commuting limits.
The most straight forward proof is to differentiate under the integral.
Newton quotients can be used to effect the differentiation if desired.
The trouble with this approach is all the difficulty is in justifying the commuting limits.
Thus less straight forward proofs are often prefered.
The only thing special about circles is that the standard proof includes some explicit computations by parameterization that are easy for circles. What is important is that the function is almost constant in a neighborhood of a point.
 
  • #3
lurflurf said:
Those links are not working.
Like much of calculus this is a problem with commuting limits.
The most straight forward proof is to differentiate under the integral.
Newton quotients can be used to effect the differentiation if desired.
The trouble with this approach is all the difficulty is in justifying the commuting limits.
Thus less straight forward proofs are often prefered.
The only thing special about circles is that the standard proof includes some explicit computations by parameterization that are easy for circles. What is important is that the function is almost constant in a neighborhood of a point.


Here are the links:

http://www2.imperial.ac.uk/~bin06/M2PM3-Complex-Analysis/m2pm3examination2008.pdf

http://www2.imperial.ac.uk/~bin06/M2PM3-Complex-Analysis/m2pm3examinsoln2008.pdf

http://www2.imperial.ac.uk/~bin06/M2PM3-Complex-Analysis/m2pm3l18(11).pdf


I've been thinking about it, and it seems that we didn't need to switch to a circular contour to prove the formula for the first derivative (although we needed it for the original Cauchy integral formula).
 

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with complex numbers and their functions. It studies the properties and behavior of complex functions, which are functions that map complex numbers to other complex numbers.

2. What is Cauchy's integral formula?

Cauchy's integral formula is a fundamental theorem in complex analysis that states that if a function is analytic (differentiable) within a closed contour, then the value of the function at any point inside the contour can be calculated by integrating the function over the contour.

3. What is the residue theorem?

The residue theorem is another important theorem in complex analysis that states that the value of a complex integral around a closed contour can be calculated by summing the residues (singularities) of the function within the contour. It is a powerful tool for evaluating complex integrals.

4. How are Cauchy's integral formula and the residue theorem related?

Cauchy's integral formula and the residue theorem are closely related as they both involve complex integrals around closed contours. In fact, the residue theorem can be seen as a generalization of Cauchy's integral formula, as it applies to functions with singularities, while Cauchy's formula only applies to analytic functions.

5. What are some real-world applications of complex analysis?

Complex analysis has many applications in various fields, including physics, engineering, and economics. It is used to solve differential equations, model fluid flow, design electrical circuits, and analyze financial data, among other things. It also has important applications in signal processing, image processing, and quantum mechanics.

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