help me figure out what type of singularity $\displaystyle z*e^{1/z}*e^{-1/z^2}$ has, please?
First, of course, the only singularity is at z= 0.
Second, $\displaystyle \frac{1}{z}- \frac{1}{z^2}= \frac{z- 1}{z^2}$ so that is the same as $\displaystyle z e^{\frac{z-1}{z^2}}$.
Lastly, $\displaystyle e^z= \sum_{n=0}^\infty \frac{z^n}{n!}$ so $\displaystyle e^{\frac{1}{z^2}}= \sum_{n=0}^\infty \frac{1}{n!}z^{-2n}$ has an infinite number of terms with negative exponent. What does that tell you?