Why is $z_0=0$ not an essential singularity of $tan(1/z)$ What is this singularity?












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Since for any $epsilon>0~exists |z|=|frac{2}{(2k+1)pi}|<epsilon$ for some $k$ large enough such that $tan(1/z)=pminfty$ there cannot be a Laurent series at $0$.



Does there need to be a Laurent series converging in a neighborhood of $0$ for it to be an essential singularity? Clearly for any $R>0$ the series does not converge in ${|z|<R}$ in this case.



What do we call this type of singularity?






complex-analysis laurent-series singularity







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  • 3




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    May be related: en.wikipedia.org/wiki/…
    $endgroup$
    – Nick
    Jan 2 at 7:45
















-1












$begingroup$


Since for any $epsilon>0~exists |z|=|frac{2}{(2k+1)pi}|<epsilon$ for some $k$ large enough such that $tan(1/z)=pminfty$ there cannot be a Laurent series at $0$.



Does there need to be a Laurent series converging in a neighborhood of $0$ for it to be an essential singularity? Clearly for any $R>0$ the series does not converge in ${|z|<R}$ in this case.



What do we call this type of singularity?






complex-analysis laurent-series singularity







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    May be related: en.wikipedia.org/wiki/…
    $endgroup$
    – Nick
    Jan 2 at 7:45














-1












-1








-1





$begingroup$


Since for any $epsilon>0~exists |z|=|frac{2}{(2k+1)pi}|<epsilon$ for some $k$ large enough such that $tan(1/z)=pminfty$ there cannot be a Laurent series at $0$.



Does there need to be a Laurent series converging in a neighborhood of $0$ for it to be an essential singularity? Clearly for any $R>0$ the series does not converge in ${|z|<R}$ in this case.



What do we call this type of singularity?






complex-analysis laurent-series singularity







share|cite|improve this question











$endgroup$




Since for any $epsilon>0~exists |z|=|frac{2}{(2k+1)pi}|<epsilon$ for some $k$ large enough such that $tan(1/z)=pminfty$ there cannot be a Laurent series at $0$.



Does there need to be a Laurent series converging in a neighborhood of $0$ for it to be an essential singularity? Clearly for any $R>0$ the series does not converge in ${|z|<R}$ in this case.



What do we call this type of singularity?






complex-analysis laurent-series singularity




complex-analysis laurent-series singularity






share|cite|improve this question















share|cite|improve this question













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edited Jan 2 at 7:43







John Cataldo

















asked Jan 2 at 7:30









John CataldoJohn Cataldo

1,1071216




1,1071216








  • 3




    $begingroup$
    May be related: en.wikipedia.org/wiki/…
    $endgroup$
    – Nick
    Jan 2 at 7:45














  • 3




    $begingroup$
    May be related: en.wikipedia.org/wiki/…
    $endgroup$
    – Nick
    Jan 2 at 7:45








3




3




$begingroup$
May be related: en.wikipedia.org/wiki/…
$endgroup$
– Nick
Jan 2 at 7:45




$begingroup$
May be related: en.wikipedia.org/wiki/…
$endgroup$
– Nick
Jan 2 at 7:45










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$0$ is not an isolated singularity. There is no disk around $0$ In which the function is analytic except for the origin.






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    $begingroup$

    $0$ is not an isolated singularity. There is no disk around $0$ In which the function is analytic except for the origin.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $0$ is not an isolated singularity. There is no disk around $0$ In which the function is analytic except for the origin.






      share|cite|improve this answer









      $endgroup$
















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        1





        $begingroup$

        $0$ is not an isolated singularity. There is no disk around $0$ In which the function is analytic except for the origin.






        share|cite|improve this answer









        $endgroup$



        $0$ is not an isolated singularity. There is no disk around $0$ In which the function is analytic except for the origin.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 8:13









        Kavi Rama MurthyKavi Rama Murthy

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