Is complex residue related to the word residue?











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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?










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    up vote
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    down vote

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    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?










    share|cite|improve this question
























      up vote
      8
      down vote

      favorite
      2









      up vote
      8
      down vote

      favorite
      2






      2





      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?










      share|cite|improve this question













      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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      asked Nov 16 at 16:20









      Benjamin Thoburn

      1579




      1579






















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          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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            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.






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              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



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






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                3 Answers
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                active

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                3 Answers
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                active

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                up vote
                6
                down vote



                accepted










                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.






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                  up vote
                  6
                  down vote



                  accepted










                  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























                    up vote
                    6
                    down vote



                    accepted







                    up vote
                    6
                    down vote



                    accepted






                    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












                    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 at 16:35









                    Nodt Greenish

                    15011




                    15011






















                        up vote
                        4
                        down vote













                        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

























                          up vote
                          4
                          down vote













                          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























                            up vote
                            4
                            down vote










                            up vote
                            4
                            down vote









                            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












                            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












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                            share|cite|improve this answer










                            answered Nov 16 at 16:28









                            José Carlos Santos

                            140k19111204




                            140k19111204






















                                up vote
                                3
                                down vote













                                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



























                                  up vote
                                  3
                                  down vote













                                  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

























                                    up vote
                                    3
                                    down vote










                                    up vote
                                    3
                                    down vote









                                    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














                                    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 at 16:38

























                                    answered Nov 16 at 16:28









                                    Masacroso

                                    12.2k41746




                                    12.2k41746






























                                         

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