when dealing with Laurent series I am confused about how to compute negative coefficients.
Laurents theorem states (in my course at least) that if f is differentiable on $D(z_0, R)/ \{z_0\}$ R>0. then $\exists a_n\in \Bbb C s.t.f(\zeta)=\sum_{n=-\infty}^{\infty}a_n(\zeta-z_0)^n$, $\zeta \in D(z_0,R)/\{z_0\}$
where $a_n=1/2\pi i\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz$
combining this with cauchy's integral formula for derivatives gives $a_n=\frac{f^n(z_0)}{n!}$.
I tried to compute the Laurent series of $\frac{1}{z^2}$ but I dont understand how to do this for $a_{-1}$ , as $a_{-1}=f^{-1}(0)/-1!$ and I'm not sure how to compute this.
Does $f^{-1}(0)$ simply mean the inverse here ?
Does $-1!=-(1!)$, i.e. does $-n!=-(n!).$