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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    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

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          The residue (latin residuere - remain) is named that way because
          $frac12pi iint_=rf(w)dw=sum_n=-infty^infty a_nint_=r(w-z_0)^n,dw= a_-1$ is what remains after integration.






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            If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminusalongrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminusa$ is a simple loop around $a$, then$$frac12pi iint_gamma f(z),mathrm dz=operatornameres_z=abigl(f(z)bigr).$$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



              $$oint z^-k, dz=begincases0,& kinBbb Zsetminus1\2pi i,& k=1endcases$$



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



              $$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$






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                The residue (latin residuere - remain) is named that way because
                $frac12pi iint_=rf(w)dw=sum_n=-infty^infty a_nint_=r(w-z_0)^n,dw= a_-1$ is what remains after integration.






                share|cite|improve this answer








                New contributor




                Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





















                  up vote
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                  The residue (latin residuere - remain) is named that way because
                  $frac12pi iint_=rf(w)dw=sum_n=-infty^infty a_nint_=r(w-z_0)^n,dw= a_-1$ is what remains after integration.






                  share|cite|improve this answer








                  New contributor




                  Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.



















                    up vote
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                    up vote
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                    down vote









                    The residue (latin residuere - remain) is named that way because
                    $frac12pi iint_=rf(w)dw=sum_n=-infty^infty a_nint_=r(w-z_0)^n,dw= a_-1$ is what remains after integration.






                    share|cite|improve this answer








                    New contributor




                    Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    The residue (latin residuere - remain) is named that way because
                    $frac12pi iint_=rf(w)dw=sum_n=-infty^infty a_nint_=r(w-z_0)^n,dw= a_-1$ is what remains after integration.







                    share|cite|improve this answer








                    New contributor




                    Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



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                    answered Nov 16 at 16:35









                    Nodt Greenish

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                        up vote
                        4
                        down vote













                        If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminusalongrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminusa$ is a simple loop around $a$, then$$frac12pi iint_gamma f(z),mathrm dz=operatornameres_z=abigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.






                        share|cite|improve this answer
























                          up vote
                          4
                          down vote













                          If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminusalongrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminusa$ is a simple loop around $a$, then$$frac12pi iint_gamma f(z),mathrm dz=operatornameres_z=abigl(f(z)bigr).$$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 $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminusalongrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminusa$ is a simple loop around $a$, then$$frac12pi iint_gamma f(z),mathrm dz=operatornameres_z=abigl(f(z)bigr).$$So, the residue is what's leftover after integrating along such a loop.






                            share|cite|improve this answer












                            If $ainmathbb C$, $rin(0,infty)$, $fcolon B_r(a)setminusalongrightarrowmathbb C$ is an analytic function, and $gammacolon[a,b]longrightarrow B_r(a)setminusa$ is a simple loop around $a$, then$$frac12pi iint_gamma f(z),mathrm dz=operatornameres_z=abigl(f(z)bigr).$$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 at 16:28









                            José Carlos Santos

                            139k18111203




                            139k18111203




















                                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



                                $$oint z^-k, dz=begincases0,& kinBbb Zsetminus1\2pi i,& k=1endcases$$



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



                                $$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$






                                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



                                  $$oint z^-k, dz=begincases0,& kinBbb Zsetminus1\2pi i,& k=1endcases$$



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



                                  $$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$






                                  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



                                    $$oint z^-k, dz=begincases0,& kinBbb Zsetminus1\2pi i,& k=1endcases$$



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



                                    $$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$






                                    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



                                    $$oint z^-k, dz=begincases0,& kinBbb Zsetminus1\2pi i,& k=1endcases$$



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



                                    $$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$







                                    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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