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Well, I have no idea: Does $\tan(1/z)$ have a Laurent series convergent on $0<|z|?

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    Where does it have poles?2012-06-10

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In short, the answer is no. The function $\tan(1/z)$ has poles at $z=\frac{1}{\pi/2+n\pi}$, which means that it is not analytic on the annulas you mentioned for any $R$.