I have question about Residues

[stgermaine]

Homework Statement

http://people.math.gatech.edu/~cain/winter99/ch10.pdf

In this PDF, I don't understand the equation at the bottom of the first page describint c sub (-1). On what basis is that equation correct? I don't know why but my diff eq teacher's giving out a test and he says there's going to be something related to the residue theorem. I haven't taken complex analysis and it's very confusing. I'd appreciate it if someone can help.

Thank you

 

[aralbrec]

http://people.math.gatech.edu/~cain/winter99/ch10.pdf

In this PDF, I don't understand the equation at the bottom of the first page describint c sub (-1). On what basis is that equation correct?

The information you are missing is the Cauchy-Goursat Theorem.

It says if you take the integral of a complex function around a closed path in the complex plane and the complex function is everywhere analytic in the region inside the path, the result is zero (this means no singularities).

Usually what follows this is a demonstration of ∫_c zⁿ dz because the result is useful. If n>=0 then everywhere enclosed by the path is analytic and the integral will be zero. Here zⁿ for n<0 has a singularity at the origin. If you integrate along a closed path that does not include the origin on its inside, then the result is still zero because zⁿ is still analytic inside the closed curve. But if the closed curve encircles the origin (eg the unit circle is the path) the theorem cannot be used.

Instead a direct integration is done. Suppose the path is the unit circle enclosing zⁿ where n<0. Then you can do a polar substitution z=r e^iθ and perform a real integration with respect to θ. If you do that, you will find the integral is still zero except when n=-1, for which you get the result:

∫_c zⁿ dz = 2∏i if n=-1 and 0 otherwise

Another theorem says paths can be deformed as long as they don't cross singularities. So this unit circle path can be deformed to any shape around the origin and you get the same result in the integral.... back to your page. You are supposing the complex function f(z) can be represented in a Laurent series. Now integrate that around a closed path enclosing z_o. The only integral term of that series that is not zero is the term belonging to (z-z_o)^-1. And the result of the integral will be 2∏i c_-1

If you are familiar with complex numbers, I can recommend chapters 3 and 4 of "Complex Variables with Applications 2nd Ed" by David Wunsch to get the ideas if this is going to be an important part of your course.