Proving $e^{ – \frac{1}{x}}$ is not holomorphic

Question

In a diffierent question I asked, in the discussion of the answer it was said to me that $$ f(z) =\begin{cases}
e^{- \frac{1}{z}} & |z|> 0 \\
0 & |z|=0
\end{cases} $$
is holomorphic only at the origin and to prove that it is not holomorphic at a point $z_0 \neq 0$ one has to approach it radially and tangentially to show the limits do not agree. I struggle to understand what was said to me and how I can algebraically do it. Does this mean that one of the limits has to be taken along the complex line passing through the origin and $z_o$ and the other has to be taken along the tangent to the circle centred at the origin of radius $|z_0|$? And if yes, how can I possibly approach this algebraically? I tried but I struggled to isolate the real and imaginary parts. If anybody can help with the algebra and approach in this, I would greatly appreciate it

Answer 1

We have that $|f( \frac{1}{it})|=1$ for all $t \in \mathbb R.$ Hence $f$ is not continuous at $0$. Consequence: $f$ is not complex differentiable at $0$.

In each $z \ne 0 $ the function $f$ is complex differentiable.