classfiying singularities \begin{equation}
h(z)=\frac{z^2e^{\frac{1}{z^2+1}}}{\sin(z^2)}
\end{equation}
 It seems to me the function has essential singularity at $z=\mp i$  
It is clear that $e^{\frac{1}{z}}$ has essential singularity at $z=0$. It can be seen that from Laurent expansion but i am not sure for essential singularities for $h(z)$
$z=0 , \lim_{z\to 0}h(z)=1$ so it is removable singularity. 
How can we see essential singularities at $z=\mp i$ ?
Thank u for your help.


We consider the Laurent series expansion of $h(z)$ with center $z=i$.
  \begin{align*}
h(z)&=\frac{z^2}{\sin(z^2)}\exp\left(\frac{1}{z^2+1}\right)\\
&=\frac{z^2}{\sin(z^2)}\exp\left(\frac{1}{2i}\cdot\frac{1}{z-i}-\frac{1}{2i}\cdot\frac{1}{z+i}\right)\tag{1}\\
&=h_1(z)\exp\left(\frac{1}{2i}\cdot\frac{1}{z-i}\right)\tag{2}\\
&=h_1(z)\sum_{n=0}^\infty\left(\frac{1}{2i}\right)^n\frac{1}{n!}(z-i)^{-n}\tag{3}
\end{align*}

Comment:


*

*In (1) we use a partial fraction decomposition to isolate the relevant part.

*In (2) we set $h_1(z)$ to
\begin{align*}
h_1(z):=\frac{z^2}{\sin(z^2)}\exp\left(-\frac{1}{2i}\cdot\frac{1}{z+i}\right)
\end{align*}
which is analytic in a neighborhood of $z=i$.

*In (3) we see that $z=i$ is an essential singularity of $\exp\left(\frac{1}{2i}\cdot\frac{1}{z-i}\right)$ and so of $h(z)$.
In the same way we can show that $z=-i$ is an essential singularity of $h(z)$.
