Determine the residue of $frac{tan z}{z}$ at $z_0=0$












0












$begingroup$


I'm having a problem finding the residue of $f(z)= dfrac{tan z}{z}= dfrac{h(z)}{z-z_0}$ at $z_0=0$ since $tan z=0$ at $z_0=0$ hence I cannot apply a proposition as it requires $h(z)$ to be analytic and nonzero at $z_0.$ I'm not also sure if I can use the Laurent Series for this.



Hints will suffice. Thank you so much.






complex-analysis







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  • 1




    $begingroup$
    I'm not sure if the function has a residue. Looks more like a removable singularity to me.
    $endgroup$
    – Ya G
    Dec 1 '18 at 4:28
















0












$begingroup$


I'm having a problem finding the residue of $f(z)= dfrac{tan z}{z}= dfrac{h(z)}{z-z_0}$ at $z_0=0$ since $tan z=0$ at $z_0=0$ hence I cannot apply a proposition as it requires $h(z)$ to be analytic and nonzero at $z_0.$ I'm not also sure if I can use the Laurent Series for this.



Hints will suffice. Thank you so much.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'm not sure if the function has a residue. Looks more like a removable singularity to me.
    $endgroup$
    – Ya G
    Dec 1 '18 at 4:28














0












0








0





$begingroup$


I'm having a problem finding the residue of $f(z)= dfrac{tan z}{z}= dfrac{h(z)}{z-z_0}$ at $z_0=0$ since $tan z=0$ at $z_0=0$ hence I cannot apply a proposition as it requires $h(z)$ to be analytic and nonzero at $z_0.$ I'm not also sure if I can use the Laurent Series for this.



Hints will suffice. Thank you so much.






complex-analysis







share|cite|improve this question











$endgroup$




I'm having a problem finding the residue of $f(z)= dfrac{tan z}{z}= dfrac{h(z)}{z-z_0}$ at $z_0=0$ since $tan z=0$ at $z_0=0$ hence I cannot apply a proposition as it requires $h(z)$ to be analytic and nonzero at $z_0.$ I'm not also sure if I can use the Laurent Series for this.



Hints will suffice. Thank you so much.






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 18:21









Did

247k23223459




247k23223459










asked Dec 1 '18 at 2:24









Mashed PotatoMashed Potato

916




916








  • 1




    $begingroup$
    I'm not sure if the function has a residue. Looks more like a removable singularity to me.
    $endgroup$
    – Ya G
    Dec 1 '18 at 4:28














  • 1




    $begingroup$
    I'm not sure if the function has a residue. Looks more like a removable singularity to me.
    $endgroup$
    – Ya G
    Dec 1 '18 at 4:28








1




1




$begingroup$
I'm not sure if the function has a residue. Looks more like a removable singularity to me.
$endgroup$
– Ya G
Dec 1 '18 at 4:28




$begingroup$
I'm not sure if the function has a residue. Looks more like a removable singularity to me.
$endgroup$
– Ya G
Dec 1 '18 at 4:28










3 Answers
3






active

oldest

votes


















1












$begingroup$

Another way to tackle problem which contains function with known taylor series ...(Or atleast we can calculate there taylor series expansion)



$tan z=z+z^3/3....$



Residue is the coefficent of $1/z$



SO $dfrac{tan z}{z}=1+z^2/3$



SO it does not have residue



Note : If don't know taylor series expansion then also you can construct then easily.



I suppose you know taylor series exapansion of $sin z $ and $cos z$



$dfrac{sin z}{cos z}=dfrac{z-z^3/6}{1-z^2/2}$



Use binomial expansion for denomiantor and do some small calculation you get required






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Here's a hint: what's the residue of $f(z)=z^n$ at $z_0=0$ for different integers $n$? Directly integrating one circle is easy.



    Now $tan(x)$ is meromorphic and so has a Taylor series around $z_0=0$, so bring it in and go on from there.



    In particular, what does $h(z_0)=0$ meromorphic around $z_0$ imply about the residue of $f(z)=h(z)/z$ at $z_0$?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't know how to answer the third question
      $endgroup$
      – Mashed Potato
      Dec 1 '18 at 2:52










    • $begingroup$
      Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
      $endgroup$
      – obscurans
      Dec 1 '18 at 2:55



















    0












    $begingroup$

    $tan z$ is analytic around $z=0$ in the region $0le |z|<frac{pi}{2}$, so it will converge in that region to its Taylor series:



    begin{align}tan z=z+frac{z^3}{3}+frac{2}{15}z^5+cdotsend{align}



    $frac{1}{z}$ is analytic everywhere except at $z=0$.



    Therefore, in the region $0<|z|<frac{pi}{2}$:



    begin{align}frac{tan z}{z}&=frac{1}{z}left(z+frac{z^3}{3}+frac{2}{15}z^5+cdotsright)\&=1+frac{z^2}{3}+frac{2}{15}z^4+cdotsend{align}



    which you can regard as a Laurent series around $z=0$ with all its coefficients $a_{-n}=0$, including the residue $a_{-1}$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Another way to tackle problem which contains function with known taylor series ...(Or atleast we can calculate there taylor series expansion)



      $tan z=z+z^3/3....$



      Residue is the coefficent of $1/z$



      SO $dfrac{tan z}{z}=1+z^2/3$



      SO it does not have residue



      Note : If don't know taylor series expansion then also you can construct then easily.



      I suppose you know taylor series exapansion of $sin z $ and $cos z$



      $dfrac{sin z}{cos z}=dfrac{z-z^3/6}{1-z^2/2}$



      Use binomial expansion for denomiantor and do some small calculation you get required






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Another way to tackle problem which contains function with known taylor series ...(Or atleast we can calculate there taylor series expansion)



        $tan z=z+z^3/3....$



        Residue is the coefficent of $1/z$



        SO $dfrac{tan z}{z}=1+z^2/3$



        SO it does not have residue



        Note : If don't know taylor series expansion then also you can construct then easily.



        I suppose you know taylor series exapansion of $sin z $ and $cos z$



        $dfrac{sin z}{cos z}=dfrac{z-z^3/6}{1-z^2/2}$



        Use binomial expansion for denomiantor and do some small calculation you get required






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Another way to tackle problem which contains function with known taylor series ...(Or atleast we can calculate there taylor series expansion)



          $tan z=z+z^3/3....$



          Residue is the coefficent of $1/z$



          SO $dfrac{tan z}{z}=1+z^2/3$



          SO it does not have residue



          Note : If don't know taylor series expansion then also you can construct then easily.



          I suppose you know taylor series exapansion of $sin z $ and $cos z$



          $dfrac{sin z}{cos z}=dfrac{z-z^3/6}{1-z^2/2}$



          Use binomial expansion for denomiantor and do some small calculation you get required






          share|cite|improve this answer









          $endgroup$



          Another way to tackle problem which contains function with known taylor series ...(Or atleast we can calculate there taylor series expansion)



          $tan z=z+z^3/3....$



          Residue is the coefficent of $1/z$



          SO $dfrac{tan z}{z}=1+z^2/3$



          SO it does not have residue



          Note : If don't know taylor series expansion then also you can construct then easily.



          I suppose you know taylor series exapansion of $sin z $ and $cos z$



          $dfrac{sin z}{cos z}=dfrac{z-z^3/6}{1-z^2/2}$



          Use binomial expansion for denomiantor and do some small calculation you get required







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 13:23









          SRJSRJ

          1,6321520




          1,6321520























              0












              $begingroup$

              Here's a hint: what's the residue of $f(z)=z^n$ at $z_0=0$ for different integers $n$? Directly integrating one circle is easy.



              Now $tan(x)$ is meromorphic and so has a Taylor series around $z_0=0$, so bring it in and go on from there.



              In particular, what does $h(z_0)=0$ meromorphic around $z_0$ imply about the residue of $f(z)=h(z)/z$ at $z_0$?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I don't know how to answer the third question
                $endgroup$
                – Mashed Potato
                Dec 1 '18 at 2:52










              • $begingroup$
                Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
                $endgroup$
                – obscurans
                Dec 1 '18 at 2:55
















              0












              $begingroup$

              Here's a hint: what's the residue of $f(z)=z^n$ at $z_0=0$ for different integers $n$? Directly integrating one circle is easy.



              Now $tan(x)$ is meromorphic and so has a Taylor series around $z_0=0$, so bring it in and go on from there.



              In particular, what does $h(z_0)=0$ meromorphic around $z_0$ imply about the residue of $f(z)=h(z)/z$ at $z_0$?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I don't know how to answer the third question
                $endgroup$
                – Mashed Potato
                Dec 1 '18 at 2:52










              • $begingroup$
                Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
                $endgroup$
                – obscurans
                Dec 1 '18 at 2:55














              0












              0








              0





              $begingroup$

              Here's a hint: what's the residue of $f(z)=z^n$ at $z_0=0$ for different integers $n$? Directly integrating one circle is easy.



              Now $tan(x)$ is meromorphic and so has a Taylor series around $z_0=0$, so bring it in and go on from there.



              In particular, what does $h(z_0)=0$ meromorphic around $z_0$ imply about the residue of $f(z)=h(z)/z$ at $z_0$?






              share|cite|improve this answer











              $endgroup$



              Here's a hint: what's the residue of $f(z)=z^n$ at $z_0=0$ for different integers $n$? Directly integrating one circle is easy.



              Now $tan(x)$ is meromorphic and so has a Taylor series around $z_0=0$, so bring it in and go on from there.



              In particular, what does $h(z_0)=0$ meromorphic around $z_0$ imply about the residue of $f(z)=h(z)/z$ at $z_0$?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 1 '18 at 2:54

























              answered Dec 1 '18 at 2:40









              obscuransobscurans

              1,027311




              1,027311












              • $begingroup$
                I don't know how to answer the third question
                $endgroup$
                – Mashed Potato
                Dec 1 '18 at 2:52










              • $begingroup$
                Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
                $endgroup$
                – obscurans
                Dec 1 '18 at 2:55


















              • $begingroup$
                I don't know how to answer the third question
                $endgroup$
                – Mashed Potato
                Dec 1 '18 at 2:52










              • $begingroup$
                Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
                $endgroup$
                – obscurans
                Dec 1 '18 at 2:55
















              $begingroup$
              I don't know how to answer the third question
              $endgroup$
              – Mashed Potato
              Dec 1 '18 at 2:52




              $begingroup$
              I don't know how to answer the third question
              $endgroup$
              – Mashed Potato
              Dec 1 '18 at 2:52












              $begingroup$
              Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
              $endgroup$
              – obscurans
              Dec 1 '18 at 2:55




              $begingroup$
              Sorry this was unclear. If $h(z_0)=0$ is meromorphic around $z_0$ then using the results for monomials tells you something specific about the residue of $h(z)/z$ at $z_0$
              $endgroup$
              – obscurans
              Dec 1 '18 at 2:55











              0












              $begingroup$

              $tan z$ is analytic around $z=0$ in the region $0le |z|<frac{pi}{2}$, so it will converge in that region to its Taylor series:



              begin{align}tan z=z+frac{z^3}{3}+frac{2}{15}z^5+cdotsend{align}



              $frac{1}{z}$ is analytic everywhere except at $z=0$.



              Therefore, in the region $0<|z|<frac{pi}{2}$:



              begin{align}frac{tan z}{z}&=frac{1}{z}left(z+frac{z^3}{3}+frac{2}{15}z^5+cdotsright)\&=1+frac{z^2}{3}+frac{2}{15}z^4+cdotsend{align}



              which you can regard as a Laurent series around $z=0$ with all its coefficients $a_{-n}=0$, including the residue $a_{-1}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $tan z$ is analytic around $z=0$ in the region $0le |z|<frac{pi}{2}$, so it will converge in that region to its Taylor series:



                begin{align}tan z=z+frac{z^3}{3}+frac{2}{15}z^5+cdotsend{align}



                $frac{1}{z}$ is analytic everywhere except at $z=0$.



                Therefore, in the region $0<|z|<frac{pi}{2}$:



                begin{align}frac{tan z}{z}&=frac{1}{z}left(z+frac{z^3}{3}+frac{2}{15}z^5+cdotsright)\&=1+frac{z^2}{3}+frac{2}{15}z^4+cdotsend{align}



                which you can regard as a Laurent series around $z=0$ with all its coefficients $a_{-n}=0$, including the residue $a_{-1}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $tan z$ is analytic around $z=0$ in the region $0le |z|<frac{pi}{2}$, so it will converge in that region to its Taylor series:



                  begin{align}tan z=z+frac{z^3}{3}+frac{2}{15}z^5+cdotsend{align}



                  $frac{1}{z}$ is analytic everywhere except at $z=0$.



                  Therefore, in the region $0<|z|<frac{pi}{2}$:



                  begin{align}frac{tan z}{z}&=frac{1}{z}left(z+frac{z^3}{3}+frac{2}{15}z^5+cdotsright)\&=1+frac{z^2}{3}+frac{2}{15}z^4+cdotsend{align}



                  which you can regard as a Laurent series around $z=0$ with all its coefficients $a_{-n}=0$, including the residue $a_{-1}$.






                  share|cite|improve this answer









                  $endgroup$



                  $tan z$ is analytic around $z=0$ in the region $0le |z|<frac{pi}{2}$, so it will converge in that region to its Taylor series:



                  begin{align}tan z=z+frac{z^3}{3}+frac{2}{15}z^5+cdotsend{align}



                  $frac{1}{z}$ is analytic everywhere except at $z=0$.



                  Therefore, in the region $0<|z|<frac{pi}{2}$:



                  begin{align}frac{tan z}{z}&=frac{1}{z}left(z+frac{z^3}{3}+frac{2}{15}z^5+cdotsright)\&=1+frac{z^2}{3}+frac{2}{15}z^4+cdotsend{align}



                  which you can regard as a Laurent series around $z=0$ with all its coefficients $a_{-n}=0$, including the residue $a_{-1}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 18:12







                  user621367





































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