Question

" Laurent Series "

Expand:

`f(z) = \frac{ {e}^{ {z}^(2) } }{ {z}^(3) }`
​

Answer 1

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

`\displaystyle e^z = \sum_(n=0)^\infty (z^n)/(n!)`

Then

`\displaystyle (e^(z^2))/(z^3) = \frac1{z^3} \sum_(n=0)^\infty (z^(2n))/(n!) = \boxed{\sum_(n=0)^\infty (z^(2n-3))/(n!)}`