Find the Laurent series for $\frac{\cos z}{z^2}$ centered at $z=0$

Question

1st attempt:

The power series expansion for $\cos z$ is: $$f(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}...$$

Dividing by $z^2$ gives:

$$f(z) = \frac{1}{z^2}-\frac{1}{2!}+\frac{z^2}{4!}-\frac{z^4}{6!}+...$$

$$f(z) = \frac{1}{z^2}+\sum_{n=0}^\infty \frac{(-1)^{n+1}z^{2n}}{(2n+2)!}$$

But this is where I get stuck.

2nd attempt:

Using the power series representation for $\cos z$, we can just represent this all as:

$$\frac{\sum_{n=0}^\infty \frac{(-1)^nz^{2n}}{(2n)!}}{z^2}$$

But then how does this relate to finding the Laurent series?

Answer 1

$$\begin{align} \frac1{z^2}\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}&=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n-2}}{(2n)!}\\\\ &=\sum_{n=-1}^{\infty}\frac{(-1)^{n-1}z^{2n}}{(2n+2)!}\\\\ &=\frac1{z^2}+\sum_{n=0}^{\infty}\frac{(-1)^{n-1}z^{2n}}{(2n+2)!} \end{align}$$

Answer 2

You are only one step away from the correct answer. Your first attempt got the sign wrong. The second one is correct, though. The following is based on your second attempt: $$ \frac{1}{z^2} \sum_{n=0}^\infty \frac{(-1)^nz^{2n}}{(2n)!} = \sum_{n=-1}^\infty \frac{(-1)^{n+1}z^{2n}}{(2n+2)!} $$ It can be broken down to the pricipal part and holomorphic part: $$ = \frac{1}{z^2} + \sum_{n=0}^\infty \frac{(-1)^{n+1}z^{2n}}{(2n+2)!} $$ Don't forget that Laurent series is unique.