# why there isn't laurent series at a non isolated singularity

I am studying the laurent series expansion around singularities.

I don't understand why at a non-isolated singularity the laurent series expansion doesn't exist, whereas we define the laurent series in an annulus. why can't we consider an annulus that avoid the non isolated singularity ?

• $\sum_{n=1}^{\infty}\frac{z^{2^k}}{k^2}$ converges on the unit disc and is singular at each point of the boundary. This is a Laurent series centered at $z=0$ without singular part, or a Laurent series centered at $z=\infty$ in which everything is singular part. Therefore, you can consider $\sum_{n=1}^{\infty}\frac{z^{-2^k}}{k^2}$ a Laurent series centered at $z=0$, in which everything is singular part. This series converges on the outside of the open unit disc, and all points of the boundary, and trivially all points inside the unit disc, are singular. – user551819 May 9 '18 at 21:14
• It is not about not associating Laurent series to non-isolated singularties, but that isolated singularities do have a Laurent series, and one that converges in a punctured neighborhood of the singularity. Of course, if the singularity is not isolated, then you can't have a Laurent series with this additional property. Observe how the series in the example has a hole the size of a disc, and not just a point. – user551819 May 9 '18 at 21:16
• A Laurent series presupposes an isolated singularity. If you examine the proof of the existence of Laurent series, you will see why. – zhw. May 9 '18 at 21:32
• @zhw. Not true. You only need a holomorphic function in an annulus. – user551819 May 9 '18 at 21:36
• In essence what I said is that you are right. Any function analytic on an annulus can be expanded in Laurent series. That is what is proven in almost every complex analysis book. My example, is just a function that is analytic in an annulus but that cannot be extended analytically beyond. You can take any such example that you like. – user551819 May 9 '18 at 22:09