Classification of singularities of a complex function

Question

I'm trying to classify in $\mathbb C\cup\{\infty\}$, the singularities of the function $$f(z)=\frac{e^{z^2}\sin(1/z)}{z}.$$ Clearly the function has only one singularity in $z=0$ and in order to tudy its nature I have tried to find the Laurent expansion of the function at $z=0$.
$$f(z)=\dfrac{1}{z}\cdot e^{z^2}\cdot\sin(1/z)=\dfrac{1}{z}\cdot\sum_{m=0}^{\infty}\frac{1}{m!}z^{2m}\cdot\sum_{n=0}^{\infty}\frac{(-)^n}{(2n+1)!}(1/z)^{2n+1}=\\ 1/z\cdot\sum_{k\in\mathbb Z}\left(\sum_{m,n\ge 0,2m-(2n-1)=k}\frac{(-)^n}{m!(2n+1)!} \right)z^k$$ and from this point I don't know how to substitute the index $n$ in function of $k$ and $m$ because I'd obtain a rational expression...

Answer 1

First hint: since $g(z)=e^{z^2}$ is entire and never vanishes, the singularity of $f$ at $0$ has the same type as $$ f_1(z)=\dfrac{f(z)}{g(z)}=\frac{1}{z}\sin\frac{1}{z} $$ that you should be able to classify.

Second hint: the function has also a singularity at $\infty$.