My idea is find the Laurent Series first but I have no idea to find it. So after receive some hints, I first have $\sin z = z(1 - \frac{z^2}{3!} +\ldots)$. Then by geometry series, I have $f(z) = \frac{1}{z^3} \frac{1}{1 - \frac{z^2}{3!} +\ldots} = \frac{1}{z^3}(1 + (\frac{z^2}{3!} +\ldots) + \ldots)$ . Then I have found that Res$(f(z),z=0) = \frac{1}{3!} = \frac{1}{6}$. It is the way to find the Residue?
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1Are you sure in the Laurent series you wrote? It says that $f(0)=0$ – Egor Ivanov Jan 08 '24 at 12:22
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@EgorIvanov Sorry I did a mistake, is the Laurent series correct right now? – Roti Canai Jan 08 '24 at 12:26
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How did you get it? – Egor Ivanov Jan 08 '24 at 12:28
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@EgorIvanov I use $sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - ... $ then $\frac{1}{sin(z)} = \frac{1}{z(1 - \frac{z^2}{3!} + \frac{z^4}{5!} - ... } = \frac{1}{z\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n}}{(2n+1)!}}$.Then multiply $\frac{1}{z^3}$ we have that. – Roti Canai Jan 08 '24 at 12:32
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I am afraid you supposed that $(\sin z)^{-1}=\sin (z^{-1})$, that is not true :( – Egor Ivanov Jan 08 '24 at 12:32
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1Something similar: https://math.stackexchange.com/questions/1266716/residue-fracezz3-sinz?rq=1 – geetha290krm Jan 08 '24 at 12:34
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@EgorIvanov I try to use another way to find it, but is the way to compute Laurent series is correct? – Roti Canai Jan 08 '24 at 13:04
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@geetha290krm I got it and the answer is $\frac{1}{6}$, thanks – Roti Canai Jan 08 '24 at 13:07
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Here you can see how to calculate correctly the Laurent series of $\frac{1}{\sin(z)}$: https://math.stackexchange.com/questions/648923/calculate-laurent-series-for-1-sinz Then it should be pretty straightforward :) – user773458 Jan 08 '24 at 13:14
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@user773458 ok thanks – Roti Canai Jan 09 '24 at 06:17