Finding the residues of $frac{cos z -1}{(e^z-1)^2}$.












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$begingroup$


I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)



I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.



Is there a better way to do this?






complex-analysis limits analysis residue-calculus holomorphic-functions







share|cite|improve this question









$endgroup$

















    -2












    $begingroup$


    I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)



    I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.



    Is there a better way to do this?






    complex-analysis limits analysis residue-calculus holomorphic-functions







    share|cite|improve this question









    $endgroup$















      -2












      -2








      -2





      $begingroup$


      I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)



      I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.



      Is there a better way to do this?






      complex-analysis limits analysis residue-calculus holomorphic-functions







      share|cite|improve this question









      $endgroup$




      I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)



      I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.



      Is there a better way to do this?






      complex-analysis limits analysis residue-calculus holomorphic-functions




      complex-analysis limits analysis residue-calculus holomorphic-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 24 '18 at 16:50









      pejetpejet

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          2 Answers
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          Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.



            $(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:



            begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}



            with $g_k(z_k)=g'_k(z_k)=1.$



            The residue of the function at $z=z_k$ is:



            begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
            \&=1-cos (2kpi i)-sin (2kpi i)
            \&=1-cosh (2kpi)-isinh (2kpi).end{align}






            share|cite|improve this answer











            $endgroup$













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              2 Answers
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              $begingroup$

              Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.






                  share|cite|improve this answer











                  $endgroup$



                  Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.







                  share|cite|improve this answer














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                  share|cite|improve this answer








                  edited Nov 26 '18 at 8:54

























                  answered Nov 24 '18 at 17:27









                  José Carlos SantosJosé Carlos Santos

                  154k22123226




                  154k22123226























                      0












                      $begingroup$

                      You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.



                      $(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:



                      begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}



                      with $g_k(z_k)=g'_k(z_k)=1.$



                      The residue of the function at $z=z_k$ is:



                      begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
                      \&=1-cos (2kpi i)-sin (2kpi i)
                      \&=1-cosh (2kpi)-isinh (2kpi).end{align}






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.



                        $(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:



                        begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}



                        with $g_k(z_k)=g'_k(z_k)=1.$



                        The residue of the function at $z=z_k$ is:



                        begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
                        \&=1-cos (2kpi i)-sin (2kpi i)
                        \&=1-cosh (2kpi)-isinh (2kpi).end{align}






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.



                          $(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:



                          begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}



                          with $g_k(z_k)=g'_k(z_k)=1.$



                          The residue of the function at $z=z_k$ is:



                          begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
                          \&=1-cos (2kpi i)-sin (2kpi i)
                          \&=1-cosh (2kpi)-isinh (2kpi).end{align}






                          share|cite|improve this answer











                          $endgroup$



                          You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.



                          $(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:



                          begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}



                          with $g_k(z_k)=g'_k(z_k)=1.$



                          The residue of the function at $z=z_k$ is:



                          begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
                          \&=1-cos (2kpi i)-sin (2kpi i)
                          \&=1-cosh (2kpi)-isinh (2kpi).end{align}







                          share|cite|improve this answer














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                          edited Dec 2 '18 at 16:20

























                          answered Dec 2 '18 at 16:11







                          user621367





































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