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I know there are several questions about Laurent series expansion by here. But I really couldn't find the expansion of

$$f(z)=\frac{1}{z\sin z} \ \ \ \mbox{and} \ \ \ g(z)=\frac{e^z}{z(1-e^{-z})}$$

around $z=0$ using the related answers. I'm able to deal only with rational functions or functions with $\sin$, $\cos$, $\log$ in the numerator. Then I'd like to ask some hint. Thanks in advance!

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\begin{align} \dfrac{1}{z\sin z} &= \dfrac{1}{z\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\right)} \\ &= \dfrac{1}{z^2}\dfrac{1}{1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots} \\ &= \dfrac{1}{z^2}\left(1+\frac{z^2}{3!}+\frac{7z^4}{3\times5!}+\cdots\right) \\ &= \dfrac{1}{z^2}+\frac{1}{3!}+\frac{7z^2}{3\times5!}+\cdots \\ \dfrac{e^z}{z(1-e^{-z})} &= \dfrac{1}{z}\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)\dfrac{1}{z-\frac{z^2}{2!}+\frac{z^3}{3!}-\cdots} \\ &= \dfrac{1}{z}\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right) \left(\dfrac1z+\dfrac12+\frac{z}{12}-\frac{z^3}{6!}+\cdots\right) \\ &= \dfrac{1}{z} \left(\dfrac1z+\dfrac32+\frac{13z}{12}+\frac{z^2}{2}+\cdots\right) \\ &= \dfrac{1}{z^2}+\frac{3}{2z}+\frac{13}{12}+\frac{z}{2}+\cdots \end{align}

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