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:


where $a_n\neq 0$.


Then: $\frac{1}{a_n+a_{n+1}(z-z_0)+...}=c_{-n}+c_{-n+1}(z-z_0)+...$ where $c_{-n}=\frac{1}{a_n}$




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!


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.