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I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $\text{tan}(1/z),\ \text{log}(z),\text{or even}\ \frac{1}{\sin(\frac{\pi}{z})}$ have a non-isolated singularity point?

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  • $\begingroup$ yes! but $e^{1/z}$ has essential singularity at $z=0$ as you expand it in laurent series expansion. number of negative terms are infinitely many $\endgroup$ – Marso Mar 7 '13 at 4:49
  • $\begingroup$ @CityOfGod I havent yet heard of essential singularity, but it isnt far removed from this section. I am going to look it up now. $\endgroup$ – Q.matin Mar 7 '13 at 4:54
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    $\begingroup$ Branch points are examples of non-isolated singularity points. Not only is there no punctured neighborhood of the branch point in which a function can be made analytic, there is no punctured neighborhood of the branch point in which a function can be made continuous! $\endgroup$ – Cameron Buie Mar 7 '13 at 5:22
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$\tan(1/z)$ has a non-isolated singularity at $z=0$, which is the limit of the singularities at $\dfrac{2}{\pi}, \dfrac{2}{3\pi}, \dfrac{2}{5\pi}, \ldots$.

The singularity of $\log(z)$ at $z=0$ is a branch point: this is on a curve where any particular branch of $\log(z)$ is discontinuous (e.g. the negative real axis in the case of the principal branch).

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A non-isolated singularity is where you can not have any further neighborhood without encountering another singularity

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