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Let $\;f: \mathbb{C} \rightarrow \mathbb{C}$ be the function

$$f(z) = \left\{ \begin{array}{cl} 0 & z = 0 \\ e^{-\frac{1}{z^2}} & z \neq 0 \end{array} \right.$$

Show that $f$ is not an entire function, but holomorphic for $\mathbb{C} \setminus \{0\}$.

I really don't get the definitions of holomorphic and entire functions (yet). Can you please tell me how this can be proved?

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    $\begingroup$ A function $f$ is holomorphic at a point $x$ if $f$ is differentiable (as a complex function) on a neighbourhood of $x$, i.e. on some small disc around $x$. $f$ is called entire if it's holomorphic at every point in the whole complex plane. Your function is holomorphic everywhere except at $0$, so it's not entire. $\endgroup$
    – fgp
    Apr 26, 2014 at 13:26

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$f$ is not even continuous at $0$: $$f(i/n)=e^{n^2}\to\infty\ne 0 $$

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  • $\begingroup$ I thought about your answer and do understand that $f$ is not continous at $0$, but why does this lead to one of the two assumptions for $f$? And what do you mean by $e^{n^2} \rightarrow \infty \neq 0$? $\endgroup$
    – muffel
    Apr 28, 2014 at 7:15
  • $\begingroup$ Every entire function is continuous everywhere.. $\endgroup$
    – user135041
    Apr 28, 2014 at 20:28
  • $\begingroup$ @muffel As Herbert said, a function that is not even continuous cannot be entire. To show discontinuity I picked the specific sequence $\frac in$, $n\in\mathbb N$ that converges to $0$, so for continuous $f$ we would have $f(i/n)\to 0$ as $n\to\infty$. however, $f(i/n)=e^{n^2}$, which clearly tends to $\infty$ instaed. $\endgroup$
    – Hagen von Eitzen
    Apr 28, 2014 at 21:31
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It is pretty easy to show that $f(z)$ is holomorphic on $\mathbb{C}\backslash\{0\}$ (e.g. by noticing that it is a composition of the exponential and a rational function), to show that it is not entire (i.e. holomorphic on the whole $\mathbb{C}$), notice (and by that, I mean prove) that it's derivative at $0$ cannot be defined (take the limit approaching $0$ in various directions).

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    $\begingroup$ The derivatives at $0$ do not exist. If you approach $0$ along the imaginary axis, the function blows up. $\endgroup$
    – Daniel Fischer
    Apr 26, 2014 at 13:39
  • $\begingroup$ @DanielFischer Ah, you're right of course. Well, that works as a proof, too. I will correct my answer. $\endgroup$
    – Daniel Robert-Nicoud
    Apr 26, 2014 at 14:08
  • $\begingroup$ @DanielRobert-Nicoud great, thank you! $\endgroup$
    – muffel
    Apr 26, 2014 at 14:51

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