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.
 A: for $z\cdot e^{\frac{-1}{z^2}}$, note if you approach to the origin along the imaginary line, say $z=ih$, we will get $ihe^{\frac{-1}{(i)^2h}}=ihe^{\frac{1}{h}}$, this obviously does not tends to zero as $h \to 0$
A: $$ze^{1/z}e^{-1/z^2}=z\left(1+\frac{1}{z}+\frac{1}{2!z^2}+...\right)\left(1-\frac{1}{z^2}+\frac{1}{2!z^4}-...\right)$$
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?
A: First, notice that 
$$\lim_{z \to 0} e^{1/z}$$ does not exist as you get different values when you approach $0$ along the real line $x + 0i$ from the right and from the left.
From there, it is not difficult to show that $\lim_{z \to 0} z e^{1/z} e^{-1/x^2}$ does not exist either.  Finally, we need to show that $\lim_{z \to 0}\frac{1}{f(z)}$ does not exist in order for $f(z)$ to have an essential singularity at $z = 0$.
In other words, you need to examine
$$
\lim_{z \to 0} \frac{e^{1/z^2}}{ze^{1/z}}.
$$
I'll leave this part to you.
