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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. –  JavaMan Oct 29 '12 at 17:11
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3 Answers

up vote 4 down vote accepted

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. –  Sam Oct 29 '12 at 21:25
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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. –  Sam Oct 29 '12 at 21:30
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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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As JavaMan and Don mentioned, I made several mistakes. Let me carefully study your replies. Also, Thank Tao; I am also trying to understand it. –  Sam Oct 29 '12 at 17:28
    
Thank JavaMan for your helpful explanation. –  Sam Oct 29 '12 at 21:31
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