Let $z_0$ be a polo singularity, f(z) is analytic in the neighbourhood excluding $z_0$.
Then $\phi(z)=\frac{1}{f(z)}$ which implies $\lim_{z\to z_0}\phi(z)=0$
So the $\phi(z)$ has the Laurent series:
$\phi(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+...$
where $a_n\neq 0$.
$f(z)=\frac{1}{(z-z_0)^n}\frac{1}{a_n+a_{n+1}(z-z_0)+...}$
Then: $\frac{1}{a_n+a_{n+1}(z-z_0)+...}=c_{-n}+c_{-n+1}(z-z_0)+...$ where $c_{-n}=\frac{1}{a_n}$
So:
$f(z)=\frac{c_{-n}}{z-z_0}+\frac{c_{-n+1}}{(z-z_0)^{n-1}}+...+\sum_\limits{n=0}^{\infty}c_n(z-z_0)^n$
Questions:
Why does the author starts at n on the following expression $\phi(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+...$? Is $\phi(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+...$ a Taylor expansion? Why not start at 0? How can the author justify that step?
Thanks in advance!
Note that $\phi$ has a removable singularity at $z_0$. Furthermore, if we extend the domain of $\phi$ adding $z_0$ to it, then $z_0$ is a zero of $\phi$ (since $\lim_{z\to z_0}\phi(z)=0$). Let $m$ be the order of that zero. The, near $z_0$,$$\phi(z)=c_m(z-z_0)^m+c_{m+1}(z-z_0)^{m+1}+\cdots$$indeed.