Well, I have no idea: Does $\tan(1/z)$ have a Laurent series convergent on $0<|z|
Does $\tan(1/z)$ has a Laurent series convergent?
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complex-analysis
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0It is preferable to use `\tan`, `\sin`, `\log`, `\cos`, `\exp` etc in math mode instead of `tan`, `sin`, `log`, `cos`, `exp` etc – 2012-06-10
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0ok, thank you, i did not know that. – 2012-06-10
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1Where does it have poles? – 2012-06-10
1 Answers
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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 $R$.