Find the domain of analyticity of

$ f(z)= \begin{cases} \exp\left(-\dfrac{1}{z^4}\right),& \text{if } z \neq 0\\ 0, & \text{if } z = 0 \end{cases} $

My Answer: Since we know that the sum, difference, product, quotient and composition of holomorphic functions are holomorphic in open sets where they are defined.

As such, since $e^z$ is holomorphic everywhere, and $\dfrac{1}{z^4}$ is holomorphic in $\mathbb{C^*}$, we have their composition to be a function holomorphic in $\mathbb{C^*}$ as well.

However, this is an essential singularity and not a removable singularity and hence cannot be "patched up". So we can conclude that the function is not analytic at $0$.

We can see that if we take the limit of the function tending to $0$, we will see the function blow up to infinity, which means the function there cannot be salvaged?


My problem with this question is i do not know how to answer the part where the function cannot be patched up and my answer sheet mentioned something about the function being unbounded and hence not analytic at $0$. Help out a confused student here !


  • $\begingroup$ Is $f$ continuous at $z=0$? $\endgroup$
    – saz
    Aug 16 '17 at 5:36
  • $\begingroup$ No it isn't continuos at 0 $\endgroup$
    – nan
    Aug 16 '17 at 5:42
  • $\begingroup$ Can $f$ be analytic at $0$ if it is not continuous at $0$? $\endgroup$
    – saz
    Aug 16 '17 at 5:54
  • $\begingroup$ No it cannot because analytic implies continuity and thus Mom continuity implies non analytic $\endgroup$
    – nan
    Aug 16 '17 at 5:57
  • $\begingroup$ Right. So you are done. $\endgroup$
    – saz
    Aug 16 '17 at 6:22

Hint: compute the limit of $$f(t e^{i \frac{\pi}4{}})$$ when $t>0$ and $t \to 0^+$ ?

  • $\begingroup$ I do not get the rationale of computing this limit, where $z = te^{i\pi/4}$. However, this answer is also included in my answer sheet, which i do not understand. If we want to show that $f$ is not continuous at $0$, we can just use the definition of continuity? $\endgroup$
    – nan
    Aug 16 '17 at 6:27
  • 1
    $\begingroup$ A powerful way to prove that a function is not continuous at $0$ (or any other point) is to find a sequence $z_n \to 0$ such that $f(z_n)$ doesn't have a limit (or has a limit different from $f(0)$). In our case, the sequence $z_n = \frac{1}{n} e^{i\pi/4}$ suits. $\endgroup$ Aug 16 '17 at 6:33

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