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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?

Thanks in advance!

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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 $c_{-n}=\frac 1{g(z_0)}=\frac 1{a_n}$.

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  • $\begingroup$ Thanks for your answer! But why does it stop at 0? Why does the author concludes the principal part series has a finite number of terms? $\endgroup$ – Pedro Gomes Apr 21 at 14:35
  • $\begingroup$ 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 $z_0$. $\endgroup$ – lEm Apr 21 at 15:17

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