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

[gangsta316]

http://www2.imperial.ac.uk/~bin06/M2...nation2008.pdf

Solutions are here.

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

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 )
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.

 

[lurflurf]

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.

 

[gangsta316]

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).