Analytic functions and integration

[physicsjock]

I have an example I want to clearify,

Let C be a semicircle from 2 to -2 which passes through 2i and let T be a semicircle from 2 to -2 which passes through -2i.

If you take the integral of z^2 dz around both paths the is the same, as the function is analytic so the integral is independent of the path,

However if you take the integral of the conjugate of z over these to paths you get

[PLAIN]http://img6.imageshack.us/img6/7467/unledpcv.jpg

 

[susskind_leon]

$$e^{-it} e^{it} = e^0 = 1$$

 

[physicsjock]

Oh okay,

So they're both i4pi, but like, what does it mean that the integrals are the same over different paths and the function is not analytic?

 

[susskind_leon]

You're integrating the same constant from 0..pi and 0..-pi, so they are negative to each other and will cancel each other out.

 

[physicsjock]

"If f(z) is analytic in a simply connected region then the int from one point to another is independent of the path joining the two points in the domain."

So here f(z) isn't analytic, is the only reason the integrals are the same the trivial one?

Because zbar isn't analytic at all

 

[susskind_leon]

No dude, I'm trying to tell you that you made a mistake in your calculation, so the integrals are NOT the same. Check your calculation please! The first integral should give 4 pi i and the second one should give -4 pi i

 

[physicsjock]

Oh right, phew,

Thanks man

hey could i ask a small question,

For scalar functions,

is the orientation of the normal vector relavent? the tu X tv,
where h(u,v) is the parametrization, tu tv its partial derivatives
because when you take the surface integral you take f(h(u,v))||tu X tv||dudv, so the orientation really doesn't matter right?

like for vector fields its the dot product of the two so the orientation of the vector does matter

 

[susskind_leon]

Yes, that's correct, the orientation of tu x tv doesn't matter if the function is scalar. What matters, of course, is the relative orientation of tu and tv, but that's pretty obvious. You know that |tu x tv| du dv is the Jacobian determinant in case you have different coordinates (like cylindrical or spherical), right? In case you're dealing with a manifold, you have the Gramian determinant.

 

[physicsjock]

Sweet thanks a lot Leon