Proof of Cauchy Integral formula

[ognik]

Hi, looking at a proof of Cauchy Integral formula, I have (at least) one question, starting from the step below

$ \int_{{C}}^{}\frac{f(z)}{z-{z}_{0}} \,dz - \int_{{C}_{2}}\frac{f(z)}{z-{z}_{0}} \,dz = 0 $ , where $ {C}_{2}$ is the smaller path around the singularity at $ {z}_{0} $

Let $z={z}_{0} + re^{i\theta} $

Then $ \int_{{c}_{2}}^{}\frac{f(z)}{z-{z}_{0}} \,dz = \int_{{c}_{2}}^{}\frac{f({z}_{0} + re^{i\theta})}{re^{i\theta}}ire^{i\theta}\,dz $

Letting r->0 gives $ = if({z}_{0})\int_{{C}_{2}}^{} \,d\theta =2\pi i f({z}_{0}) $

I follow all that (hope I explain this well enough) - but after letting $z={z}_{0} + re^{i\theta} $, isn't this now different to the f(z) of the original contour? We let r tend to 0 for $ {C}_{2}$ , but C is some larger R?

 

[Fernando Revilla]

... but after letting $z={z}_{0} + re^{i\theta} $, isn't this now different to the f(z) of the original contour? We let r tend to 0 for $ {C}_{2}$ , but C is some larger R?

Yes, you are right, but any proof of the Cauchy's integral formula will use several and true properties. A silly but perhaps clarifying example:

If $f(x)=x,$ then $\displaystyle\int_{-2}^2f(x)\;dx=\int_{-1}^1f(x)\;dx=0,$ however $[-2,2]$ is larger than $[-1,1].$

 

[ognik]

Ah, what I was forgetting is that Cauchy's Integral theorem also shows that analytic functions within the simply connected region are path independent.

1. So all analytic functions will be conservative? This is also shown by the Cauchy-Riemann equations?

2. Random thought - are all real analytic functions conservative?

3. Please check my understanding - the 'starting step' I used, shows that it doesn't matter if the paths C and $C_{2}$ are of different lengths (radii). Therefore we can use any convenient path for the inner path, and it must be a path that allows us to remove the radius of that path from the equation?

(This is my second pass through the material, picking up on stuff that didn't sink in completely the first time :-))

 

[Fernando Revilla]

So all analytic functions will be conservative?

Yes, on simply connected regions.

This is also shown by the Cauchy-Riemann equations?

Right.

Random thought - are all real analytic functions conservative?

If I understand your question, if $f:[a,b]\to \mathbb{R}$ is analytic, then $f$ has a primitive $F,$ hence $\displaystyle\int_a^bf(x)\;dx=F(b)-F(a).$ In that sense, we can say that $f$ is conservative.

Therefore we can use any convenient path for the inner path, and it must be a path that allows us to remove the radius of that path from the equation?

It seems to me that you are only talking about a technical reason, for example the possibility of removing the radius when computing an integral. There is a deeper and theoric reason: the invariance of integration on homtopic curves.

(This is my second pass through the material, picking up on stuff that didn't sink in completely the first time :-))

Third time's a charm. :)

 

[ognik]

It seems to me that you are only talking about a technical reason, for example the possibility of removing the radius when computing an integral. There is a deeper and theoric reason: the invariance of integration on homtopic curves.

You are right and I hadn't even heard of holotropic curves, after some browsing I now have a vague idea of it and yes - that seems a big part of the justification for the technique, loosing the R is just a mathematical convenience. Thanks again.