I want to know about the singularities of the function $$f(z) = e^{-1/z^2}, \ z \neq 0, \qquad \\ 0, \quad z = 0$$. Explain in detail please.
For poles i need to look at the points where $e^{1/z^2} = 0$. But Exponential are never $0$, so no poles. Now if i expand its series expansion, then there will be infinitely many terms containing negative powers of $z$, so there must be non isolated essential singularity at $0$. Am i correct?
We have $e^z= \sum_{n=0}^{\infty}\frac{z^n}{n!}$, hence, for $z \ne 0$,
$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!z^{2n}}$.
Consequence: $f$ has at $0$ an essential singularity. You wrote "non isolated essential singularity". But the singularity at $0$ is isolated !
A generalization: if $f$ is an entire function and if $g$ is defined by $g(z)=f(1/z)$, then we have:
$g$ has at $0$ an essential singularity $ \iff f$ is not a polynomial.
Try a proof !
In the above case we have $f(z)=e^{-z^2}$