Originally Posted by

**Bruno J.** Many different questions there!

We have to be a bit picky here. For a point to qualify as a removable singularity of a function (in the usual sense), the function must first be holomorphic on a punctured open neighbourhood of this point. This is not the case for a point on a branch cut; the function is not even continuous along the cut! But we can extend the notion of a removable singularity to points which lie on a branch cut (other than the branch points); because we can, indeed, define the function properly there (by moving the branch cut, or, what is equivalent, by analytic continuation).

So I'd say that points on a branch cut (other than the branch points) qualify as removable singular points in some extended sense. They definitely do not qualify as essential singularities or poles. (An essential singularity is essentially a pole of infinite order.)

Essentially, what I'm saying is that if you look at the function $\displaystyle \log(1+x)$ as a single-valued function defined on its Riemann surface, then there are only two troublesome points, $\displaystyle -1$ and $\displaystyle \infty$, which are essential singularities. Branch cuts are only a construct which allow us to "fit" inside the plane a function which is "too big" for the plane.

It isn't! You can expand your function in a Laurent series (in fact, in a Taylor series) around every point other than $\displaystyle -1$ and $\displaystyle \infty$. You just have to remember that you will not obtain a single-valued function in this way.

Yes; a function has no Laurent expansion around a nonisolated singular point (which is necessarily an essential singularity).