Laurent series of $z\mapsto\frac{z^4}{z^2-1}$ in $\infty$

Question

I want to calculate the Laurent series and the main part in $\infty$ of $$f:z\mapsto\frac{z^4}{z^2-1}.$$

The Taylor series of $f(\frac{1}{z})$ in $0$ is $$\sum_{n=0}^{\infty}z^{2(n-1)}=1/z^2+1+z^2+z^4+...$$

What is the Laurent series in $\infty$ and what the main part in $\infty$? How do I get them?

Answer 1

As you said, do $\;z\to\frac1z\;$ and develop around zero:

$$f\left(\frac1z\right)=\frac{\frac1{z^4}}{\frac1{z^2}-1}=\frac1{z^2-z^4}=\frac1{z^2}\frac1{1-z^2}=\frac1{z^2}\left(1+z^2+z^4+\ldots\right)=...$$

Answer 2

The general approach to computing the series of $f$ around $\infty$ is to first write the series of $f(\frac 1 z)$ around $0$, and then substitute $\frac 1 z$ instead of $z$ in the result. In your case,

$$f(\frac 1 z) = \frac {\frac 1 {z^4}} {\frac 1 {z^2} - 1} = \frac {1} {z^2 - z^4} = \frac 1 {z^2} (1 + z^2 + z^4 + \dots) = \frac 1 {z^2} + 1 + z^2 + z^4 + \dots$$

and replacing now $z$ by $\frac 1 z$ you get

$$f(z) = z^2 + 1 + \frac 1 {z^2} + \frac 1 {z^4} + \dots .$$

By definition, the principal part of a Laurent series is the sum of the terms with negative powers, which in your case means $\frac 1 {z^2} + \frac 1 {z^4} + \dots$.