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This is for homework in my complex analysis class, and I think there may be a mistake. I wanted to make sure I didn't miss anything obvious before I bring it up to the professor. The problem asks to show that the function $$ f(z) = \begin{cases} e^{-\frac{1}{z^4}}, & \text{if } z \neq 0 \\ 0, & \text{if } z = 0 \end{cases} $$ is not holomorphic at $z = 0$. The definition of holomorphic that we are using is:

A function $f$ is holomorphic at $z_0$ if $\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$ exists.

For the $f$ I defined above, I found that (and WolframAlpha agrees) $$ \lim_{z \to 0} \frac{e^{-\frac{1}{z^4}}}{z} = 0, $$ implying the function is indeed holomorphic at 0. Did I miss anything obvious, or should I bring this to the professor's attention?

tylerc0816
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  • Was Wolfram, and were you, assuming that $z$ is real when this limit was found? – Michael Hardy Sep 04 '13 at 16:38
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    Gah! I knew I was doing something stupid. In the future, is there a way to make wolfram take the limit in the complex sense? Or do I have to convert it into a limit in $\mathbb{R}^2$? – tylerc0816 Sep 04 '13 at 16:41
  • I wouldn't be surprised if that can be done, but I don't know how to do it. – Michael Hardy Sep 04 '13 at 16:47
  • For fun, don't forget to check if Cauchy--Riemann's equations are satsified at $z=0$! (See also http://math.stackexchange.com/questions/291233) – mrf Sep 04 '13 at 21:06

2 Answers2

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Hint: $$z_n=\frac{1}{n}e^{i\pi/4} \to 0, \quad n \to \infty.$$

njguliyev
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The limit $$ \lim_{z\to0} f(z)=\lim_{z\to 0}\frac{e^{-1/z^4}}{z} $$ Certainly does exist if you consider this limit for $z\in\mathbb R$. However, things get a little weirder when you look at the neighborhood of $0$ in $\mathbb C$.

For example, we can take the limit along the path $z=re^{\pi i/4}=\frac{1+i}{\sqrt2}r$, where $r\in\mathbb R$. We find that $$ \lim_{r\to0}f(z(r))=\lim_{r\to 0}\frac{e^{-1/(re^{\pi i/4})^4}}{r}=\lim_{r\to0}\frac{e^{1/r^4}}{r} $$ Where $r\in\mathbb R$. If $f$ is indeed holomorphic, you should get $0$. What do you actually get?

Ben Grossmann
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