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$z\cdot e^{1/z}\cdot e^{-1/z^2}$ 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.

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    You seem to have made two mistakes in your post. First, $\lim_{z \to 0} z e^{-1/z^2}$ is not zero. Second, you seemed to have lost the $e^{1/z}$ term, so you chould check your algebra.2012-10-29

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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 $h \to 0$

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    I think Tao's answer is easiest for me to understand. Thank you very much.2012-10-29
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$ze^{1/z}e^{-1/z^2}=z\left(1+\frac{1}{z}+\frac{1}{2!z^2}+...\right)\left(1-\frac{1}{z^2}+\frac{1}{2!z^4}-...\right)$

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?

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    Thank you, Don, but I have not figured out how to use the series expansion to determine the singularity.2012-10-29
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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.

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    Thank JavaMan for your helpful explanation.2012-10-29