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 Ch. 7.2- Contour Integration ?s 
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Joined: 20 Aug 2009, 22:54
Posts: 2
Post Ch. 7.2- Contour Integration ?s
Was wondering if anyone could shed some light on why contour integrals "work," so to speak. From what the book says, a homologous deformation results in a closed loop that encircles the origin (or undefined point in the domain) that is not concerned with point a or b. Does this mean that such a deformation will obtain a single, unique result no matter what interval on the complex function you wish to integrate, or will results differ by a constant (radius of closed loop?) but differentiate to the same derivative?

Thanks in advance.


20 Aug 2009, 23:26

Joined: 13 Aug 2009, 00:08
Posts: 13
Post Re: Ch. 7.2- Contour Integration ?s
I'll have a go at answering this: The contour integral is defined as the difference between the value of the function at the end points, so:
\int_a^b f^\prime(z) dz = f(b) - f(a)
where {df}/{dz}=f^\prime.

Let's consider this for f(z) = 1/z, as Penrose does. For any closed contour the end points a and b are coincident but for a contour around the break point (the singularity at the origin) their arguments differ by 2\pi, so the answers differ by 2{\pi}i, so f(b) - f(a) = 2{\pi}i.

Penrose's discussion on p126 in chapter 7.2 provides one way of looking at this for closed contours -- the important results are for closed contours.

To address your questions:

(1) The deformation idea is about changing the path to other paths using holomorphic transformations -- for example in the case of f(z)=1/z (RTR page 126), a closed contour around the origin will give a different result for the contour integral than will a closed contour that does not include the origin. In fact, the latter will always give the answer zero because any two end-points chosen will have the same argument; only a closed contour around the origin will have a different argument.

(2) Results will differ by a constant -- in fact, by 2{\pi}i in this case -- for the reasons given. The path of the contour does not matter (amazingly!) provided it is closed and has the singularity inside it. If the singularity is outside the closed contour then the result of the integral is zero. If the singularity is inside then the result is 2{\pi}i provided the closed contour goes around the singularity once in the anti-clockwise direction (the answer is different if you go around more than once or in a different direction).

(3) Not sure about your final question fragment "will results differ by a constant (radius of closed loop?) but differentiate to the same derivative?". As mentioned above, the radius of the closed loop is irrelevant. Also, the answer to the integration is an actual number (usually complex) -- it's a definite integral -- this means that you cannot differentiate it again and get back to where you started.

My discussion here is not at all rigorous, and there are other things to consider when the singularities have higher orders, but I hope it helps you to see what is going on.


22 Aug 2009, 16:17

Joined: 20 Aug 2009, 22:54
Posts: 2
Post Re: Ch. 7.2- Contour Integration ?s
Ahhh, I got it now. Thanks for that explanation--it is much appreciated!!


24 Aug 2009, 03:24

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Ch. 7.2- Contour Integration ?s
Has anyone figured out why x+iy to r=Sqrt (x**2 + y**2), Theta = x/y, is a one to many mapping yet? Look people, this whole attempt to fit the observable world with present day mathematics is an exercise in masturbation!


24 Sep 2009, 07:07

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Ch. 7.2- Contour Integration ?s
The reason that all mathematicians give for their abstract thinking is this :

MY MIND CAN MODEL THE WORLD!

So a Baryon here, a Hooligan there,,we can model this world,,so spacktheth Johnnie!


24 Sep 2009, 07:19

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Ch. 7.2- Contour Integration ?s
Oh yes,,even the Great God Gauss was invovled with the fact that equations gave rise to complex numbers. After all, at the end of your day your yard measures x*y = -1

THEN WHAT?
NEW MATH?
No, but the multipication thing must go back to addition.


24 Sep 2009, 07:28

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Ch. 7.2- Contour Integration ?s
BTW it is the "MULTIPLICATION" thing that brings in the "PRIME NUMBERS" , with addition there are no such numbers.


24 Sep 2009, 07:39
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Ch. 7.2- Contour Integration ?s
harried wrote:
Has anyone figured out why x+iy to r=Sqrt (x**2 + y**2), Theta = x/y, is a one to many mapping yet?


If you mean x+iy to r=\sqrt (x^2 + y^2), ~~\theta = \tan^{-1} (y/x)

then the answer is yes. It's because \tan^{-1} is a multi-valued function.


24 Sep 2009, 09:54

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Ch. 7.2- Contour Integration ?s
I had a teacher who said "If you can't get an answer to your problem add another dimension, and fit your answer in"!
x+iy --> r*e**(i*Theta) gives you that fool's freedom!
Is electron spin really a periodic human invention?


02 Nov 2009, 07:35
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Joined: 07 Jun 2008, 08:21
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Post Re: Ch. 7.2- Contour Integration ?s
Harried
What are you on?
I'm sorry, but I can't make any sense of your ramblings. Can anyone else?
Vasco


02 Nov 2009, 16:44
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