# Understanding the Laurent series $\frac{1}{a_n+a_{n+1}(z-z_0)+…}=c_{-n}+c_{-n+1}(z-z_0)+…$

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:

$$f(z)=\frac{c_{-n}}{z-z_0}+\frac{c_{-n+1}}{(z-z_0)^{n-1}}+...+\sum_\limits{n=0}^{\infty}c_n(z-z_0)^n$$

Question:

How does the author derive the expression $$\frac{1}{a_n+a_{n+1}(z-z_0)+...}=c_{-n}+c_{-n+1}(z-z_0)+...$$? How does the author find out $$c_{-n}=\frac{1}{a_n}$$? What kind of technique is being used on this step?

The function $$g(z)=a_n+a_{n+1}(z-z_0)+...$$ is a Taylor series, with $$a_n\neq 0$$, by continuity there is an open ball around $$z_0$$ so that $$g(z)$$ does not vanish on the ball. So $$\frac 1{g(z)}$$ is holomorphic in that ball and can be expanded into Taylor series. This series is $$c_{-n}+c_{-n+1}(z-z_0)+...$$ in your question. And we see that $$c_{-n}=\frac 1{g(z_0)}=\frac 1{a_n}$$.
• Which one are you referring to? If you are asking about $c_{-n}$, it is because of the fact that $\frac 1{g(z)}$ is holomorphic near $z_0$. – lEm Apr 21 at 15:17