# the type of singularity

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• Apr 19th 2010, 02:34 AM
Archi
the type of singularity
help me figure out what type of singularity $z*e^{1/z}*e^{-1/z^2}$ has, please?
• Apr 19th 2010, 04:15 AM
HallsofIvy
First, of course, the only singularity is at z= 0.

Second, $\frac{1}{z}- \frac{1}{z^2}= \frac{z- 1}{z^2}$ so that is the same as $z e^{\frac{z-1}{z^2}}$.

Lastly, $e^z= \sum_{n=0}^\infty \frac{z^n}{n!}$ so $e^{\frac{1}{z^2}}= \sum_{n=0}^\infty \frac{1}{n!}z^{-2n}$ has an infinite number of terms with negative exponent. What does that tell you?