Questions regarding the Residue theorem (complex analysis)

Question

I am trying to figure out how to calculate the residue of a function f(z) at z=$\infty$. My best guess is to define $g(z)=f(\frac{1}{z})$ then find the Laurent series for g(z) at z=0 and then find the coefficient for$\frac{1}{z}$. Is this correct?

Another question I have is if it's possible for the residue to be equal to zero?

A simple example I tried to solve was: Find al the isolated singular points and calculate the residue at each point for: $f(z)=\frac{e^{3z}}{z-2}$

Ill appreciate any help!

Answer 1

1. No, though getting at the correct formulation of it really requires some notion of Riemann surfaces (or similar constructions). In brief, the issue is that although we can define the residue at $z = \infty$ by considering the residue of $f(\phi^{-1})$ at $z = \phi(\infty)$ for some suitable map $\phi$, we also want it to be independent of $\phi$; at the very least, there's no canonical choice of it. (We also need to define what it means for $\phi$ to be holomorphic at $\infty$, etc.) In a bit more detail, we can consider in a closed, simple curve $\gamma$ in the Riemann sphere $\hat{\mathbb{C}}$ that doesn't pass through $\infty\in \hat{\mathbb{C}}$. Then $\gamma$ splits $\hat{\mathbb{C}}$ into two connected components, and we want the residue theorem to hold on each (being careful about the orientation of $\gamma$ in either case). You can check that your definition of the residue at $\infty$ doesn't work for, say, $f(z) = z$.

2. Sure: The function $f(z) = 1/z^2$ has a (double) pole at $z = 0$ with a residue of $0$.

3. For the reasons outlined in part 1, this sort of question only deals with residues for $z\in \mathbb{C}$. The function $f(z) = e^{3z}/(z - 2)$ is meromorphic on $\mathbb{C}$ with its only pole at $z = 2$, and computing the residue there is a straightforward application of the residue calculus.