Happy spring break.
I feel like I get it. But I'm not quite there... Thanks so much!
I really appreciate your help, but please forgive me; my book gave very little motivation or background on residues. I have a vague idea of what they are, I'm not really sure where we're supposed to find them, etc.
Also, I'm not sure why $\displaystyle \frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$. If anything, wouldn't it be equal to $\displaystyle \frac{1}{(z-z_0)}\sum_{n=0}^{\infty}a_n$. How does this relate to residues?
Here's an exercise that might help to motivate "residue".
Integrate $\displaystyle \oint (z- z_0)^n dz$ around a circle with center at $\displaystyle z_0$ and radius R where n can be any integer. Do that by writing $\displaystyle z= Re^{i\theta}+ z_0$ and integrating with respect to $\displaystyle \theta$.
You should see that the integral is 0 for all n except n= -1, and, in that case, is $\displaystyle 2\pi i$.
Now, suppose f(z) has a "pole of order n" at $\displaystyle z_0$. That means that f(z) is not analytic, so cannot be written as a Taylor's series about $\displaystyle z_0$ 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 $\displaystyle z_0$" means that $\displaystyle (z- z_0)^n f(z)$ is analytic at $\displaystyle z_0$ but not for any lower power of $\displaystyle z- z_0$. That, in turn, means that we can write $\displaystyle (z- z_0)^nf(z)$ as a Taylor's series about $\displaystyle z_0$: $\displaystyle (z- z_0)^n f(z)= a_0+ a_1(z-z_0)+ a_2(z- z_0)^2+ a_3(z- z_0)^3+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n(z- z_0)^n+ \cdot\cdot\cdot$.
Dividing both sides by $\displaystyle (z-z_0)^n$, $\displaystyle f(z)= a_0(z- z_0)^{-n}+ a_1(z- z_0)^{-n+1}+ a_2(z- z_0)^{-n+2}+ a_3(z- z_0)^{-n+3}+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n+ \cdot\cdot\cdot$.
Now integrate the right side term by term around a path with $\displaystyle z_0$ 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:
$\displaystyle \oint f(z)dz= a_{n-1}(2\pi i)$. That number is the "residue of f(z) at $\displaystyle z= z_0$".