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

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It is preferable to use \tan, \sin, \log, \cos, \exp etc in math mode instead of tan, sin, log, cos, exp etc –  user17762 Jun 10 '12 at 5:35
    
ok, thank you, i did not know that. –  Bunuelian Trick Jun 10 '12 at 5:39
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Where does it have poles? –  WimC Jun 10 '12 at 6:08
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up vote 5 down vote accepted

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 $R$.

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