Is complex residue related to the word residue?












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$begingroup$


I know little formal math terminology and don't understand much of anything about complex analysis. Also, if this isn't a good starting point for complex integration feel free to say (I'm learning about it partly for Cauchy's residue theorem).



My first and intuitive idea of residue has to do with remainder, subtraction, division, etc., But more generally something that's leftover, extra, or unused by an operation or something. Do these ideas tie together?






complex-analysis







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    8












    $begingroup$


    I know little formal math terminology and don't understand much of anything about complex analysis. Also, if this isn't a good starting point for complex integration feel free to say (I'm learning about it partly for Cauchy's residue theorem).



    My first and intuitive idea of residue has to do with remainder, subtraction, division, etc., But more generally something that's leftover, extra, or unused by an operation or something. Do these ideas tie together?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      8












      8








      8


      2



      $begingroup$


      I know little formal math terminology and don't understand much of anything about complex analysis. Also, if this isn't a good starting point for complex integration feel free to say (I'm learning about it partly for Cauchy's residue theorem).



      My first and intuitive idea of residue has to do with remainder, subtraction, division, etc., But more generally something that's leftover, extra, or unused by an operation or something. Do these ideas tie together?






      complex-analysis







      share|cite|improve this question









      $endgroup$




      I know little formal math terminology and don't understand much of anything about complex analysis. Also, if this isn't a good starting point for complex integration feel free to say (I'm learning about it partly for Cauchy's residue theorem).



      My first and intuitive idea of residue has to do with remainder, subtraction, division, etc., But more generally something that's leftover, extra, or unused by an operation or something. Do these ideas tie together?






      complex-analysis




      complex-analysis






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      asked Nov 16 '18 at 16:20









      Benjamin ThoburnBenjamin Thoburn

      350212




      350212






















          3 Answers
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          6












          $begingroup$

          The residue (latin residuere - remain) is named that way because
          $frac{1}{2pi i}int_{|w-z_0|=r}f(w)dw=sum_{n=-infty}^{infty} a_nint_{|w-z_0|=r}(w-z_0)^n,dw= a_{-1}$ is what remains after integration.






          share|cite|improve this answer









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            4












            $begingroup$

            If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminus{a}longrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminus{a}$ is a simple loop around $a$, then$$frac1{2pi i}int_gamma f(z),mathrm dz=operatorname{res}_{z=a}bigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.






            share|cite|improve this answer









            $endgroup$





















              3












              $begingroup$

              Yes, it is. The residue of the Laurent series of a function in an annulus is the coefficient $c_{-1}$ of this series, that is what is left after integration of the function in a loop on the annulus (that is, after the integration of the Laurent series that represent the function in this annulus).



              Note that



              $$oint z^{-k}, dz=begin{cases}0,& kinBbb Zsetminus{1}\2pi i,& k=1end{cases}$$



              Hence (assuming a Laurent series around a pole in the origin for simplicity) we have that



              $$oint f(z), dz=ointsum_{kinBbb Z}c_k z^k, dz=sum_{kinBbb Z}c_k oint z^k, dz=c_{-1}2pi i$$






              share|cite|improve this answer











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                Your Answer





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                3 Answers
                3






                active

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                3 Answers
                3






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                active

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                6












                $begingroup$

                The residue (latin residuere - remain) is named that way because
                $frac{1}{2pi i}int_{|w-z_0|=r}f(w)dw=sum_{n=-infty}^{infty} a_nint_{|w-z_0|=r}(w-z_0)^n,dw= a_{-1}$ is what remains after integration.






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  The residue (latin residuere - remain) is named that way because
                  $frac{1}{2pi i}int_{|w-z_0|=r}f(w)dw=sum_{n=-infty}^{infty} a_nint_{|w-z_0|=r}(w-z_0)^n,dw= a_{-1}$ is what remains after integration.






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    The residue (latin residuere - remain) is named that way because
                    $frac{1}{2pi i}int_{|w-z_0|=r}f(w)dw=sum_{n=-infty}^{infty} a_nint_{|w-z_0|=r}(w-z_0)^n,dw= a_{-1}$ is what remains after integration.






                    share|cite|improve this answer









                    $endgroup$



                    The residue (latin residuere - remain) is named that way because
                    $frac{1}{2pi i}int_{|w-z_0|=r}f(w)dw=sum_{n=-infty}^{infty} a_nint_{|w-z_0|=r}(w-z_0)^n,dw= a_{-1}$ is what remains after integration.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 16 '18 at 16:35









                    Nodt GreenishNodt Greenish

                    29813




                    29813























                        4












                        $begingroup$

                        If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminus{a}longrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminus{a}$ is a simple loop around $a$, then$$frac1{2pi i}int_gamma f(z),mathrm dz=operatorname{res}_{z=a}bigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminus{a}longrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminus{a}$ is a simple loop around $a$, then$$frac1{2pi i}int_gamma f(z),mathrm dz=operatorname{res}_{z=a}bigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminus{a}longrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminus{a}$ is a simple loop around $a$, then$$frac1{2pi i}int_gamma f(z),mathrm dz=operatorname{res}_{z=a}bigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.






                            share|cite|improve this answer









                            $endgroup$



                            If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminus{a}longrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminus{a}$ is a simple loop around $a$, then$$frac1{2pi i}int_gamma f(z),mathrm dz=operatorname{res}_{z=a}bigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 16 '18 at 16:28









                            José Carlos SantosJosé Carlos Santos

                            158k22126228




                            158k22126228























                                3












                                $begingroup$

                                Yes, it is. The residue of the Laurent series of a function in an annulus is the coefficient $c_{-1}$ of this series, that is what is left after integration of the function in a loop on the annulus (that is, after the integration of the Laurent series that represent the function in this annulus).



                                Note that



                                $$oint z^{-k}, dz=begin{cases}0,& kinBbb Zsetminus{1}\2pi i,& k=1end{cases}$$



                                Hence (assuming a Laurent series around a pole in the origin for simplicity) we have that



                                $$oint f(z), dz=ointsum_{kinBbb Z}c_k z^k, dz=sum_{kinBbb Z}c_k oint z^k, dz=c_{-1}2pi i$$






                                share|cite|improve this answer











                                $endgroup$


















                                  3












                                  $begingroup$

                                  Yes, it is. The residue of the Laurent series of a function in an annulus is the coefficient $c_{-1}$ of this series, that is what is left after integration of the function in a loop on the annulus (that is, after the integration of the Laurent series that represent the function in this annulus).



                                  Note that



                                  $$oint z^{-k}, dz=begin{cases}0,& kinBbb Zsetminus{1}\2pi i,& k=1end{cases}$$



                                  Hence (assuming a Laurent series around a pole in the origin for simplicity) we have that



                                  $$oint f(z), dz=ointsum_{kinBbb Z}c_k z^k, dz=sum_{kinBbb Z}c_k oint z^k, dz=c_{-1}2pi i$$






                                  share|cite|improve this answer











                                  $endgroup$
















                                    3












                                    3








                                    3





                                    $begingroup$

                                    Yes, it is. The residue of the Laurent series of a function in an annulus is the coefficient $c_{-1}$ of this series, that is what is left after integration of the function in a loop on the annulus (that is, after the integration of the Laurent series that represent the function in this annulus).



                                    Note that



                                    $$oint z^{-k}, dz=begin{cases}0,& kinBbb Zsetminus{1}\2pi i,& k=1end{cases}$$



                                    Hence (assuming a Laurent series around a pole in the origin for simplicity) we have that



                                    $$oint f(z), dz=ointsum_{kinBbb Z}c_k z^k, dz=sum_{kinBbb Z}c_k oint z^k, dz=c_{-1}2pi i$$






                                    share|cite|improve this answer











                                    $endgroup$



                                    Yes, it is. The residue of the Laurent series of a function in an annulus is the coefficient $c_{-1}$ of this series, that is what is left after integration of the function in a loop on the annulus (that is, after the integration of the Laurent series that represent the function in this annulus).



                                    Note that



                                    $$oint z^{-k}, dz=begin{cases}0,& kinBbb Zsetminus{1}\2pi i,& k=1end{cases}$$



                                    Hence (assuming a Laurent series around a pole in the origin for simplicity) we have that



                                    $$oint f(z), dz=ointsum_{kinBbb Z}c_k z^k, dz=sum_{kinBbb Z}c_k oint z^k, dz=c_{-1}2pi i$$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 16 '18 at 16:38

























                                    answered Nov 16 '18 at 16:28









                                    MasacrosoMasacroso

                                    13.1k41746




                                    13.1k41746






























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