0
$\begingroup$

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 !

**

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

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.