Consider \(f(z) = \mathrm{cosec}(\tfrac{\pi}{z})\). Then \(f(z)\) has a pole at \(z=\tfrac{1}{n}\) for any nonzero integer \(n\), which makes the singularity at \(0\) non-isolated.
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Mark Hennings
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3 years, 2 months ago

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@Mark Hennings
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Thnxs dude can you think of one more function
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R G
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3 years, 2 months ago

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Generalize my previous example. Let \(f(z)\) be any nonconstant entire function with an infinite number of zeros. Then the zeros are isolated, countable and unbounded. Then \(f(z^{-1})^{-1}\) has a non-isolated singularity at \(z=0\), since \(w^{-1}\) is a pole for any zero \(w\neq0\) of \(f(z)\). For example, \(f(z) = e^z - 1\) would do the trick.
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Mark Hennings
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3 years, 2 months ago

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TopNewestConsider \(f(z) = \mathrm{cosec}(\tfrac{\pi}{z})\). Then \(f(z)\) has a pole at \(z=\tfrac{1}{n}\) for any nonzero integer \(n\), which makes the singularity at \(0\) non-isolated. – Mark Hennings · 3 years, 2 months ago

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– R G · 3 years, 2 months ago

Thnxs dude can you think of one more functionLog in to reply

Generalize my previous example. Let \(f(z)\) be any nonconstant entire function with an infinite number of zeros. Then the zeros are isolated, countable and unbounded. Then \(f(z^{-1})^{-1}\) has a non-isolated singularity at \(z=0\), since \(w^{-1}\) is a pole for any zero \(w\neq0\) of \(f(z)\). For example, \(f(z) = e^z - 1\) would do the trick. – Mark Hennings · 3 years, 2 months ago

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