# the type of singularity

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