Question

" Laurent Series "

Expand:
$f(z) = \frac{ {e}^{ {z}^{2} } }{ {z}^{3} }$
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Answer 1

Recall that for all $z\in\Bbb C$,

$\displaystyle e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$

Then

$\displaystyle \frac{e^{z^2}}{z^3} = \frac1{z^3} \sum_{n=0}^\infty \frac{z^{2n}}{n!} = \boxed{\sum_{n=0}^\infty \frac{z^{2n-3}}{n!}}$