# Find the domain of analyticity of $f(z)= \exp\left(-\dfrac{1}{z^4}\right)$

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?

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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 !

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

Hint: compute the limit of $$f(t e^{i \frac{\pi}4{}})$$ when $t>0$ and $t \to 0^+$ ?
• 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?
• 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. Aug 16 '17 at 6:33