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?
complex-analysis
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up vote
8
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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?
complex-analysis
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
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
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
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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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
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.
New contributor
add a comment |
up vote
5
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.
New contributor
add a comment |
up vote
5
down vote
up vote
5
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.
New contributor
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.
New contributor
New contributor
answered Nov 16 at 16:35
Nodt Greenish
739
739
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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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 16 at 16:28
José Carlos Santos
139k18111203
139k18111203
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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
$$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$
add a comment |
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
$$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$
add a comment |
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
$$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$
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
$$oint f(z), dz=ointsum_kinBbb Zc_k z^k, dz=sum_kinBbb Zc_k oint z^k, dz=c_-12pi i$$
edited Nov 16 at 16:38
answered Nov 16 at 16:28
Masacroso
12.2k41746
12.2k41746
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