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Answered

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" Laurent Series "

Expand:
[tex] f(z) = \frac{ {e}^{ {z}^{2} } }{ {z}^{3} } [/tex]
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Sagot :

LammettHash

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]

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