Consider the following integral: $$\frac{1}{2\pi i }\oint_C f(z) dz$$ with $$f(z)= \frac{z^2+1}{z^3}$$ and $C$ is the unit circle centered at the origin.
Consider two ways to do it:
- Enclosing the singular points inside $C$: This is actually its residue at $z=0$ (order = $3$). So $$\text{Res}(f(z),z=0)=\frac{1}{(3-1)!}\frac{d^2}{dz^2}(z^2+1)\bigg|_{z=0}=1$$
- Enclosing the singular points outside $C$ (by including the point at infinity):
From the textbook I read, 1. and 2. should give the same answer. However, I have no idea how to do 2.?
Can anyone please let me know how to deal with 2.?
The function $f(z) = \frac{z^2+1}{z^3}$ is holomorphic at $\infty$, yet still has a residue there. There are no other singular points outsize $C$ to consider.
$$ -\frac{1}{2\pi i} \oint_{C} f(z)\,dz=\text{Res}_{z=\infty} f(z) = -\lim_{z\to \infty} zf(z)= -1.$$
Here is a source defining the residue at infinity.
http://www.stat.phys.titech.ac.jp/~yokoyama/note5.pdf