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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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 – Myshkin Mar 07 '13 at 04:49
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@CityOfGod I havent yet heard of essential singularity, but it isnt far removed from this section. I am going to look it up now. – Q.matin Mar 07 '13 at 04:54
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2Branch 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! – Cameron Buie Mar 07 '13 at 05:22
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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).
Robert Israel
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1Allegedly there is a third type of a non-isolated singularity, known as 'boundary' points. Can you give an example of a function with such a singularity? – nilo de roock Apr 09 '21 at 19:27
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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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Idk why this is downvoted so much - it captures the gist. A nonisolated singularity is one where you don't have a punctured disk that doesnt have another singularity – Max0815 Oct 01 '23 at 05:51