I've been looking for a concise explanation of how to obtain the Laurent expansion for $$\frac{1}{\sin(z)}$$ My attempt at it has me confused by it pretty quickly. I start with the knowledge that $$\sin(z)=z-\frac{1}{3!}z^3+\frac{1}{5!}z^5-...$$ Then I move that to the denominator to begin expanding the function at hand. I have gotten it to the form of a geometric series; namely, $$\frac{1}{\sin(z)}=\frac{1}{z}\cdot\frac{1}{1-(\frac{1}{3!}z^2-\frac{1}{5!}z^4+\cdots)}$$ I am unsure of how to proceed from here. Any help appreciated.
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Do you want the general term, or just the first few? – Chappers May 03 '15 at 21:38
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I would love to see how to obtain the first few. I'm using this to calculate a bunch of residues, so I do need it term-by-term. Thanks!! @Chappers – Taylor May 03 '15 at 21:43
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1So for the first few terms, let $w = \frac{z^2}{3!} - \frac{z^4}{5!} + \dots$, then expand $\frac{1}{1-w}$ into a geometric series, and in $$\frac{1}{z}(1 + w + w^2 + \dotsc),$$ expand as many powers of $w$ as far as you need. For the residue, you need not go far at all. – Daniel Fischer May 03 '15 at 22:01
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You may use polynomial long division to evaluate $\dfrac z{z-z^3/3!+z^5/5!-z^7/7!}$. – Raymond Manzoni May 03 '15 at 22:52