For this problem, I have three questions:
1. Why do we calculate the residue at the infinity? Is it because all the four poles are in contained in the disk?
2. When calculating $f(1/z)$, don't we just subsitute the z in the function with $1/z$? Why do I get a different answer?
3. Why do we need Laurent series at last? And how do we get it?
Thank you very much!
Best Answer
We say that a function $f$ has a singularity at $\infty$ if the open set $G$ where $f$ is defined contains a punctured disc centered at $\infty$, or in other words is defined outside of some circle centered at the origin with positive radius. So for example, $\frac{1}{\sin{z}}$ does not have a singularity at $\infty$ since it has singularities at $n\pi$ for integer $n$; there's no circle $\vert z \vert = R$ outside which the function is defined everywhere.
Yes, that's correct. $\infty$ is removable, essential or a pole of $f(z)$ if $0$ is a removable/essential singularity or a pole of $f(1/z)$. As to why you're getting a different answer, maybe you're making a mistake when calculating $f(1/z)$.
The residue of a function at an isolated singularity $z_0$ is the coefficient of the $(z-z_0)^{-1}$ term in that function's Laurent series expansion centered at $z_0$. You can get this by writing out what the Taylor expansion of that function centered at that singularity; there are also some simpler formulas you can use in some cases.