I read here that since $\sin(z)$ has an essential singularity at $\infty$, $1/\sin(z)$ also has an essential singularity at $\infty$. This got me thinking: Is it true that if $f(z)$ has an essential singularity at $z_0$ then $1/f(z)$ also has a singularity at $z_0$ and can this be proved in terms of Laurent series?
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Do you know the classification of isolated singularities into removable s., poles, and essential s.? – Martin R Feb 24 '16 at 14:49
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@MartinR Yes, at least I thought I did. Removable is when $\lim_{z\rightarrow z_0}(f(z))$ is finite, pole order $n$ is when $\lim_{z\rightarrow z_0}((z-z_0)^nf(z))$ is finite, and essential is when $\lim_{z\rightarrow z_0}((z-z_0)^nf(z))$ is not finite for any finite $n$. – Quantum spaghettification Feb 24 '16 at 14:52
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@mrf I grant that this question is a duplicate, in the fact that it asks if $f(z)$ has an essential singularity then does $1/f(z)$ but the part about weather it can be proved using a Laurent Series is not, I am still curious so know if this proof is possible? – Quantum spaghettification Feb 24 '16 at 15:04
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1@Quantumspaghettification Basically no. If you start with $f(z) = \sin(1/z)$, we have an essential singularity at $z=0$, but $1/f$ does not admit a Laurent series around $z=0$ (since the singularity at $z=0$ of $1/f$ is non-isolated). – mrf Feb 24 '16 at 15:25
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That $1/f(z)$ has a singularity can be shown by using Picard's theorem. We know that $f(z)$ takes all values except possibly one arbitrarily near $z_0$. This means that $f(z)$ will be arbitrarily close to $0$ arbitrarily near $z_0$ and therefor $1/f(z)$ is not bounded in any neighborhood of $z_0$.
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