$f(z)$ has essential singularity at $z=0$, what type of singularity $f(z^2+z)$ has?

$f(z)$ has essential singularity at $z=0$ so it can be written has $\sum_{n=0}^{-\infty} c_nz^n=c_0+\frac{c_{-1}}{z}+\frac{c_{-2}}{z^2}+...+$ substituting $z=z^2+z$ will leave all the power to be negative e.g $c_0+\frac{c_{-1}}{z^2+z}+\frac{c_{-2}}{(z^2+z)^2}+...+$

So $f(z^2+z)$ has essential singularity too, moreover can we even "get rid" of essential singularity?

  • 1
    $\begingroup$ Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $\infty$), but there will always remain an essential singularity "somewhere" in the plane. $\endgroup$ – Clayton Jul 16 '18 at 21:34
  • 3
    $\begingroup$ Possible duplicate of Singularities of Composition of Functions $\endgroup$ – Clayton Jul 16 '18 at 21:38
  • $\begingroup$ Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$. $\endgroup$ – GEdgar Jul 18 '18 at 14:23
  • $\begingroup$ @Clayton That question is different: it asks about $g \circ f$ where $f$ has a(n essential) singularity. This question is about $f \circ g$ (with $g(z) = z^2 + z$). $\endgroup$ – user88319 Jul 23 '18 at 13:51

Here is a rigorous proof of the fact that $g(z)\equiv f(z^{2}+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<\delta $ with $0<\delta <1$. If $|w|$ is sufficiently small take $z=\frac {-1+c} 2$ where $c$ is such that $c^{2}=1+4z$ and $\Re c>0$. Then $z^{2}+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| \to \infty $ as $z \to 0$ and an argument similar to the previous one shows that $|f(z)| \to \infty $ as $z \to 0$, again a contradiction.

| cite | improve this answer | |
  • $\begingroup$ In the last line you mean $|f|\to \infty$ $\endgroup$ – Mathmath Dec 3 '19 at 9:26
  • $\begingroup$ @Mathmath Yes thanks for pionting out. $\endgroup$ – Kavi Rama Murthy Dec 3 '19 at 9:29

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.