# what is the significance of the annulus in laurent series?

I am looking at ways which you can write a function as a series. I am aware that one can use the Taylor series. I am currently trying to understand the Laurent series. I understand there are cases where the Taylor series will not work as all the terms in it cannot have a negative degree. The Laurent series is there to aid this kind of problem.

Upon examining texts, including the wikipedia page I see that for a function $f:C\subset \mathbb{C} \rightarrow \mathbb{C}$, the Laurent series at a point $z_0 \in C$ is given by

$$f(z) = \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$$

where the $a_n$ is a somewhat a generalized version of the Cauchy integral formula. I have the following questions with regards to this:

• Why do you need to have an annulus around $z_0$
• When calculating the $a_n$'s, can you pick any closed curves surrounding $z_0$ within the annulus and integrate along their boundaries?

If $f$ is not analytic at $z_0$ (and the singularity is isolated and not removable), then the Laurent series at $z_0$ will converge in an annulus around $z_0$. Why an annulus? What follows assumes you accept why Taylor series gives you a radius/disk of convergence.
If you ignore negative powers, then it is like a regular Taylor series and you already know Taylor series have a radius of convergence (outer circle) $|z-z_0| < R$.
Now if you focus on the convergence of the negative powers, think of it as a Taylor series of a different change of variable. $\sum_{n > 0} a_{-n} (z-z_0)^{-n}$ becomes $\sum_{n>0} a_{-n} y^n$ for $y = (z-z_0)^{-1}$. Note that there is a radius of convergence for this series, $|y| < r$, which translates to convergence when $|z-z_0| > r^{-1}$.
Conclude convergence in annulus $r^{-1} < |z-z_0| < R$. This is the largest region for which the Laurent series converges (since this is the case for the individual Taylor series).