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Proof that function defined $f(0)=0$, $f(z) = e^{-z^{-4}}$ for $z \neq 0$, is holomorphic in $\mathbb{C}\setminus\{0\}$ but is not differentiable at $0$.

I have this question, and searched for an answer, and I saw these (1), (2) and (3). I use the Stein Complex Analysis book. To show that function is not differentiable at 0, I see that if I take $z=re^{\frac{i \pi}{4}}$ and apply the limit $$ \displaystyle \lim_{r \to 0} \frac{f(z)}{z} = \lim_{r \to 0} \frac{e^{-1/ (re^{\frac{i\pi}{4}})^4}}{re^{i \pi/4}}$$ so $$\lim_{r \to 0} \frac{e^{-1/ (re^{\frac{i\pi}{4}})^4}}{re^{i \pi/4}} = \lim_{r \to 0} \frac{e^{\frac{1}{r^4}}}{r}.$$

I see that $$(re^{\frac{i\pi}{4}})^4 = r^4(\cos 4\pi/4 + i \sin 4\pi/4) = -r^{4}.$$

But why $re^{i \pi/4} = r$?

Other question is, who I can proof that limit exist only using the limit properties? This is a singularity?

Thanks.

José Carlos Santos
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  • How'd you go from$$r^4(\cos(4\pi/4)+i\sin(4\pi/4))$$to$$r^{-4}$$? – Simply Beautiful Art Aug 22 '17 at 13:51
  • It is certainly true that $\lim_{z\to 0}\frac{f(z)}{z}$ does not exist. However, neither does $\lim_{z\to 0} f(z)$, although you seem to implicitly assume that the latter is $0$. Perhaps because you've done something like $$e^{-0^{-4}}=e^{-(+\infty)}=e^{-\infty}=0$$ however, there is no such thing as $\pm\infty$ in $\Bbb C$ and, actually, for any $u\in\Bbb C\setminus{0}$ the equation $\exp(z)=u$ has arbitrarly large solutions. –  Aug 22 '17 at 13:53
  • Simply Beautiful Art: Sorry, I made a mistake, is $-r^4$ not $r^{-4}$ –  Aug 22 '17 at 14:53

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