What type of singularity is this? $z\cdot e^{1/z}\cdot e^{-1/z^2}$ at $z=0$.
My answer is removable singularity. 
$$
\lim_{z\to0}\left|z\cdot e^{1/z}\cdot e^{-1/z^2}\right|=\lim_{z\to0}\left|z\cdot e^{\frac{z-1}{z^2}}\right|=\lim_{z\to0}\left|z\cdot e^{\frac{-1}{z^2}}\right|=0.
$$
But someone says it is an essential singularity. I don't know why.

$$ze^{1/z}e^{-1/z^2}=z\left(1+\frac{1}{z}+\frac{1}{2!z^2}+...\right)\left(1-\frac{1}{z^2}+\frac{1}{2!z^4}-...\right)$$
So this looks like an essential singularity, uh? 
I really don't understand how you made the following step:
$$\lim_{z\to 0}\left|z\cdot e^{\frac{z-1}{z^2}}\right|=\lim_{z\to0}\left|z\cdot e^{\frac{-1}{z^2}}\right|$$
What happened to that $\,z\,$ in the exponential's power?
