So, I'm a little confused about removable singularities. Consider the function below:
$$f(z)=\frac{1}{(1+z^2)^{2/3}}$$
Obviously, we have isolated singularities at the points $z = \pm i$. However, are they removable? In my studies, I have found that removable singularities exist if the following are true:
the limit of $(z-z_o)f(z)$ as $z$ approaches $z_o$ is zero.
the limit of $f(z)$ as $z$ approaches $z_o$ exists.
I have deduced that the first limit above is, indeed, equal to zero. So, can I deduce that the singularity is removable? I ask because wolfram alpha says that I have a pole at these points, not a removable singularity. Further, when I ask for a series about $z=i$, it has a $(z-i)^{-\frac{2}{3}}$ term. What's going on here; it seems like derived results are contradicting each other?
To summarize: before we can remove a singularity, we must isolate it first: that is, make sure that the function is holomorphic in some annulus $\{z:0<|z-z_0|<r\}$. Which isn't the case here, because starting with some branch of $f$ and traveling around $z_0$ in a circle, we observe $\arg f $ changing by $4\pi/3$.
The asymptotic behavior of $f$ near $z_0$ does not even come into play.