How to get the Laurent Series of $$\frac{e^z}{z(1+z^2)}$$ at $z=0$? I know the answer is
$$\frac{1}{z} + 1 - \frac{1}{2}z - \frac{5}{6}z^2 + \frac{13}{24}z^3 + \frac{101}{120}z^4 + O(z^5)$$
But I don't know how to do it, wanna know the step...
2026-03-25 21:54:14.1774475654
Laurent Series of $\frac{e^z}{z(1+z^2)}$ [at z=0]
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There is no simple formula for that. You know that$$\frac{e^z}{1+z^2}=\left(1+\frac z{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}\cdots\right)\left(1-z^2+z^4-z^6+\cdots\right).$$You compute the first few terms and then you divide by $z$ to get the Laurent series.