\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.


1 Answer 1


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*}


  • 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 $h(z)$.


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