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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.

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Thnxs dude can you think of one more function

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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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