4
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $z_0$.

$\endgroup$
  • $\begingroup$ Are you writing "isolated" for "essential"? $\endgroup$ – zhw. Sep 3 '17 at 17:24
  • $\begingroup$ @zhw. That's it. I've edited my answer. $\endgroup$ – José Carlos Santos Sep 3 '17 at 17:32
1
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.