TITLE: Isolated singularites of $f(z)=\frac{\log(1+z)}{z^2}$?
QUESTION [0 upvotes]: Let $f:\mathbb{C}\setminus\{0\}\to\mathbb{C}$ be the holomorphic function given by $$f(z)=\frac{\log(1+z)}{z^2}.$$
I know that $z_0=0$ is a pole of degree $1$.
My question: What about those points where the complex logarithm is not defined? E. g. $z_1=-1$. Is that also a singularity, if so, what kind of singularity is it and why?
Thank you!

REPLY [0 votes]: Well the function is differentiable near -1, so yes it has a singularity there. A holomorphic function can only have the following isolated singularities: poles, removable singularities, or neither which are so called essential singularities. It can also have a branch point, which is the case here because the function is discontinuous around a small circle centered around $z=-1$ (why?). It's more proper name is a logarithmic branch point.
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