i was studyng this function, and on wolfram alpha it says that there are no poles. But why is $z=0$ not a pole?
(sorry for my bad english)
2026-03-25 01:26:59.1774402019
Poles of $\sin(1/z)$
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It is not a pole because it is an essential singularity:$$\sin\left(\frac1z\right)=\frac1z-\frac1{3!z^3}+\frac1{5!z^5}-\frac1{7!z^7}+\cdots$$