It is clear to me that $e^{1/z}$ has an essential singularity at $z=0$, since it has infinitely many terms in Laurent Series expansion principle part.
I was quite sure about my question as well, till I found some text saying that $e^{-1/z^2}$ doesn't have a singularity at all. How could this be possible, Laurent Series should work here also. Or may be texts are wrong that I found.
Kindly help thanx and regards.
When $f$ is restricted in the real line, and we set $f(0)=0$, then $f$ is $C^\infty$, but NOT real analytic (i.e., expressible as a power series).
However, in $\mathbb C$ it DOES have an essential singularity at $z=0$.