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Math Help - the type of singularity

  1. #1
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    the type of singularity

    help me figure out what type of singularity z*e^{1/z}*e^{-1/z^2} has, please?
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  2. #2
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    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?
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