Laurent series and apparent essential singularity

Question

After reading "Mathematics for physicists" by Susan M. Lea I encountered a subtlety that I can't turn my head around (p. 128). Consider function

$$f(z)=\frac{1}{z^2-1} = \frac{1}{2}\left[\frac{1}{z-1}-\frac{1}{1+z}\right]$$

which clearly has 2 simple poles at $z=\pm 1$. If we want to compute its Laurent series around $z=1$, we would get:

$$\frac{1}{1+z}=\frac{1}{2+(z-1)} = \frac{1}{(z-1)\left(1+\frac{2}{z-1}\right)}$$ $$=\frac{1}{z-1}\sum_{n=0}^\infty\frac{(-2)^n}{(z-1)^{n}}$$ and thus: $$\frac{1}{z^2-1} = \frac{1}{2}\left[\frac{1}{z-1}-\sum_{n=0}(-1)^n\frac{2^n}{(z-1)^{n+1}}\right] = \frac{1}{2}\sum_{n=1}^\infty\frac{(-2)^n}{(z-1)^{n+1}}$$

What I see here is an infinite number of negative powers of (z-1) and thus, I would be tempted to conclude that there is an essential singularity at this point. What prevents me from concluding this as it is clear that there is no essential singularity from the analytic function?

Answer 1

As I mentioned in the comments, the series you derived does not exists in a punctured neighborhood of the point $z=1$ as it only converges in $|z-1|>2$.

You should have expanded $\frac{1}{1+z}$ via $$\frac{1}{1+z}=\frac{1}{2-(1-z)}=\frac{1}{2}\frac{1}{1-\frac{1-z}{2}}=\frac12\sum_{n=0}^\infty\left( \frac{1-z}{2}\right)^n$$ and it converges in $|z-1|<2$. So about $z=1$ your full Laurent series is $$f(z)=\frac{1}{2}\left[-\frac{1}{1-z}-\frac12\sum_{n=0}^\infty \left(\frac{1-z}{2}\right)^n \right]=-\frac{1}{4}\sum_{n=-1}^\infty \left(\frac{1-z}{2}\right)^n. $$