Is $\exp(-1/z^2)$ differentiable at $0$? Let $f: \mathbb{C} \rightarrow \mathbb{C}: \begin{cases} \exp(-1/z^2) & z \neq 0 \\ 0 & z=0 \end{cases}$ be a function. Is $f$ differentiable in $0$?
Suppose $f$ is differentiable in $a$, then $\lim_{z \rightarrow 0} \frac{f(z)-f(0)}{z-0}=\lim_{z \rightarrow 0} \frac{\exp(-1/z^2)}{z}$ has to exist in $\mathbb{C}$. I'm not sure how to evaluate this limit. Can I use L'Hopital's rule here or does that only work for real functions?

You can try and evaluate the limit where $z \to 0$ along the real line and along the imaginary line. We have
$$ \lim_{x \to 0} \frac{e^{-1/x^2}}{x} =  0 $$
while
$$ \lim_{x \to 0} \frac{e^{-1/(ix)^2}}{ix} = (-i) \cdot \lim_{x \to 0} \frac{e^{1/x^2}}{x} = \infty. $$
The first limit can be calculated using L'Hopital as a real limit or in any other way.
