Laurent-Development of zero function

Question

The following "idendity" contradicts the uniqueness of the Laurent Development: $$ 0=\frac{1}{z-1}+\frac{1}{1-z}=\frac{1}{z}\frac{1}{1-1/z}+\frac{1}{1-z}=\sum_{n=1}^{\infty}\frac{1}{z^n}+\sum_{n=0}^{\infty}z^n=\sum_{n=-\infty}^{\infty}z^n $$ 1) Where is the mistake? 2) When does $\sum_{n=-\infty}^{\infty}z^n$ converge?

My answers are:

1)

The series $\sum_{n=1}^{\infty}\frac{1}{z^n}$ converges for $\lvert z\rvert >1$.

The series $\sum_{n=0}^{\infty}z^n$ converges fpr $\lvert z\rvert <1$.

So the mistake is to develop both fractions in a geometric series, because this does not work!

2)

The negative degree power series of the Laurentseries converges for $\lvert z\rvert>1$, the positive degree power series converges for $\lvert z\rvert < 1$. So the whole Laurentseries converges for $$ \left\{z\in\mathbb{C}: 1<\lvert z\rvert <1\right\}=\emptyset. $$

So the Laurentseries never converges?

Answer 1

You are correct. The series $\delta(z) = \sum\limits_{n=-\infty}^\infty z^n$ does not converge anywhere (so there is no contradiction -- to get uniqueness for a Laurent series expansion you need convergence on an open annulus ).

Now even though this series $\delta(z)$ does not converge anywhere, it is tremendously useful in formal series manipulations. It even has a name: the formal Dirac delta. It is an example of an expansion of zero (since it would be an alternate Laurent series expansion of zero if it converged).

Expansions of zero and especially the formal delta function play a significant role in the theory of vertex algebras. For more details, James Lepowsky and Haisheng Li's text "Introduction To Vertex Operator Algebras And Their Representations" is a great reference.