Recall that for all [tex]z\in\Bbb C[/tex],
[tex]\displaystyle e^z = \sum_{n=0}^\infty \frac{z^n}{n!}[/tex]
Then
[tex]\displaystyle \frac{e^{z^2}}{z^3} = \frac1{z^3} \sum_{n=0}^\infty \frac{z^{2n}}{n!} = \boxed{\sum_{n=0}^\infty \frac{z^{2n-3}}{n!}}[/tex]
