Ways to see if a function $f(z)$ is holomorphic

Question

Despite I'm going to use an example to illustrate what I'm referring to, this pretends to be a general question about the procedure to find where a function is holomorphic.

Suppose we have

$$f(z)=\begin{cases}e^{-1/z^2} \quad z\neq 0\\ 0 \qquad\ \ \ \ z=0\end{cases}$$

We want to see if this function is holomorphic.

As far as I'm concerned, we have 2 ways to, in practise, see it:

• If this function has partial derivatives everywhere except on $0$, we can find if C-R equation holds for $f(z)$. Then, everywhere where $f(z)$ satisfies this equations, the function will be holomorphic.

• If it's hard (like on this example) to separate $f(z)$ in its real and imaginary part to verify that, we can verify if the function is $C^1(\Omega)$, and then we know that $\partial f/ \partial \bar{z}=0$ iff the function is holomorphic.

Am I correct? Did I leave any case? Is there another (easier) way to see that $f(z)$ is (isn't) holomorphic?

Answer 1

It's easier to note $e^z$ is holomorphic everywhere, and $-1/z^2$ is holomorphic on $\mathbb C\setminus\{0\}.$ Therefore the composition, $e^{-1/z^2},$ is holomorphic on $\mathbb C\setminus\{0\}.$ What happens at $0?$ There $f$ is not differentiable, or even continuous. To see this, consider $f(z)$ as $z\to 0$ along the imaginary axis.

Answer 2

Your function is not even continuous at the origin, so of course it can't be holomorphic. A way to show this: try to take $z \to 0$ on the real axis, and then on the imaginary axis. You will get different limits, so the function is not continuous.

Answer 3

The first thing to check is whether your function is continuous or not. And it isn't: $\lim_{x\to0^+}f(x+xi)\neq0$.

Besides, you should be aware of the fact that a point of the domain of $f$ may well be a solution of the Cauchy-Riemann equations and, even so, $f$ may not be differentiable there.