Happy spring break.
I feel like I get it. But I'm not quite there... Thanks so much!
Also, I'm not sure why . If anything, wouldn't it be equal to . How does this relate to residues?
Here's an exercise that might help to motivate "residue".
Integrate around a circle with center at and radius R where n can be any integer. Do that by writing and integrating with respect to .
You should see that the integral is 0 for all n except n= -1, and, in that case, is .
Now, suppose f(z) has a "pole of order n" at . That means that f(z) is not analytic, so cannot be written as a Taylor's series about but can be written as a "Laurent" series (with negative powers up to -n).
Basically, saying "f(z) has a pole of order n at " means that is analytic at but not for any lower power of . That, in turn, means that we can write as a Taylor's series about : .
Dividing both sides by , .
Now integrate the right side term by term around a path with in its interior. As before, the integral of all powers of x, except one, will be 0. The exception is for the "-1" power- we will have:
. That number is the "residue of f(z) at ".