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Question. In univariate complex analysis, are essential isolated singularities preserved under non-zero holomorphic functions?

For example, if we've already proved that $e^{1/z}$ has an essential singularity at $0$, can we deduce that $e^{e^{1/z}}$ also has one at zero, without making any computations?

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Let $f\colon D\longrightarrow\mathbb C$ be an analytic function and suppose that it has an essential singularity at some point $z_0$. Let $g$ be a non-constant entire function. Then $g\circ f$ also has an essential singularity at $z_0$. This is so because, by the Casorati-Weierstrass, if $U$ is a neighborhood of $z_0$ such that $U\setminus\{z_0\}\subset D$, then $f\bigl(U\setminus\{z_0\}\bigr)$ is a dense subset of $\mathbb C$. On the other and, it is an easy corollary of the Liouville theorem that the image of $g$ is dense. Therefore, $(g\circ f)(D)$ is dense too. And it follows from this that $g\circ f$ also has an essential singularity at $z_0$.

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  • $\begingroup$ Are you writing "isolated" for "essential"? $\endgroup$
    – zhw.
    Sep 3, 2017 at 17:24
  • $\begingroup$ @zhw. That's it. I've edited my answer. $\endgroup$ Sep 3, 2017 at 17:32
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I believe essential isolated singularities are preserved under non-constant, entire functions.

Indeed, if $g$ has an essential singularity at $0$, then Great Picard tells us that the image under $g$ of any punctured neighbourhood of $0$ is either $\mathbb C$ or $\mathbb C$ minus a point. Moreover, if $f$ is a non-constant entire function, then by applying Little Picard to $f$, we learn that the image under $f \circ g$ of any punctured neighbourhood of $0$ is $\mathbb C$ minus at most two points, and in particular, this image is dense in $\mathbb C$. Therefore, it is impossible for $f \circ g$ to have a pole or a removable singularity at $0$.

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