How to get the Laurent-series of $1/f(z)^p $

Question

Given an analytic function $f (z)=z-z^{p+1}+O(z^{2p+1} )$ how can I find the Laurent-series of $1/f(z)^p $. I can understand that $$f(z)^p=z^p-pz^{2p}+\binom {p}{2}z^{3p}+O(z^{4p})$$ However, how can I go on now. I think I can find the First coefficient via evaluating $$1/(f(z)^p)*z^p$$ at zero, giving the coefficient $1$ for $z^{-p}$. I would be happy enough to get to know how to get the Next $2,3$ nonzero coefficients. Thanks

Answer 1

Since $g(z)=\frac{z^p}{f(z)^p}$ is holomorphic in $z=0$ you can compute the taylor series of $g$ at $z=0$ and you get $$ \frac1{f(z)^p}=\frac{g(z)}{z^p} = \frac{\sum_{k=0}^\infty \frac{g^{(k)}(0)}{k!}z^k}{z^p}=\sum_{k=0}^\infty \frac{g^{(k)}(0)}{k!}z^{k-p}. $$