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Given this function $f(z) = \frac{e^z}{z^2}$

I need to determine whether there is a pole, or it is removable or it has the essential singularity.

My Approach: I know that $e^z = \sum_{n=0}^{inf}\frac{z^n}{n!}$ in power series expansion, but I cannot seem to find any removable due to lowest term is 1 from the power series expansion.

Now the question is in this case, does this mean there is a pole at 0 with order of 2, or does it have an essential singularity which implies that both pole and removable cannot happen in the following function.

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  • $\begingroup$ Have you considered $\lim_{z\to 0} z^2 f(z)$? $\endgroup$ – David P Oct 27 '19 at 22:54
  • $\begingroup$ @DavidPeterson, you mean using f(z) =$a_{n}(z - z_{0})^n$ formula? $\endgroup$ – mathrock1453 Oct 27 '19 at 22:57
  • $\begingroup$ Also I am supposed to use the power series expansion... $\endgroup$ – mathrock1453 Oct 27 '19 at 23:01
  • $\begingroup$ $\frac{1}{f(z)}=z^2e^{-z}$, which is analytic and has a zero of order two at $z=0$. Therefore, $f$ has a pole of order $2$ at $z=0$. $\endgroup$ – conditionalMethod Oct 27 '19 at 23:11
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$f(z)=\frac 1 {z^{2}}+ \frac 1 z+\frac 1 {2!}+\frac z {3!}+\cdots$ so $f$ has a pole of order $2$ at $0$.

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