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Residue theorem: understood it intuitively but cannot prove mathematically
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In my opinion, residue theorem’s underlying principle is quite beautiful: one decomposes a contour integral into infinitely many small circular contour integrals around every point, and most of them are zero, except a few that encircle singularities. Those are called residues, and add them up. This is basically how residue theorem works.
However, when I try to prove it by Green’s theorem, I face some difficulties. ($f=u+iv$)
$$oint_C f(z)dz=oint(u+iv)dx+i(u+iv)dy=iint_D fracpartial (u+iv)partial x-ifracpartial (u+iv)partial ydxdy$$
I don’t see how can this be transformed into a sum of residues/circular contour integrals around singularities.
Can you please show how?
Thanks in advance.
complex-analysis
asked Jul 25 at 3:47
Szeto
4,0931521
add a comment |Â
up vote
1
down vote
favorite
In my opinion, residue theorem’s underlying principle is quite beautiful: one decomposes a contour integral into infinitely many small circular contour integrals around every point, and most of them are zero, except a few that encircle singularities. Those are called residues, and add them up. This is basically how residue theorem works.
However, when I try to prove it by Green’s theorem, I face some difficulties. ($f=u+iv$)
$$oint_C f(z)dz=oint(u+iv)dx+i(u+iv)dy=iint_D fracpartial (u+iv)partial x-ifracpartial (u+iv)partial ydxdy$$
I don’t see how can this be transformed into a sum of residues/circular contour integrals around singularities.
Can you please show how?
Thanks in advance.
complex-analysis
asked Jul 25 at 3:47
Szeto
4,0931521
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my opinion, residue theorem’s underlying principle is quite beautiful: one decomposes a contour integral into infinitely many small circular contour integrals around every point, and most of them are zero, except a few that encircle singularities. Those are called residues, and add them up. This is basically how residue theorem works.
However, when I try to prove it by Green’s theorem, I face some difficulties. ($f=u+iv$)
$$oint_C f(z)dz=oint(u+iv)dx+i(u+iv)dy=iint_D fracpartial (u+iv)partial x-ifracpartial (u+iv)partial ydxdy$$
I don’t see how can this be transformed into a sum of residues/circular contour integrals around singularities.
Can you please show how?
Thanks in advance.
complex-analysis
asked Jul 25 at 3:47
Szeto
4,0931521
In my opinion, residue theorem’s underlying principle is quite beautiful: one decomposes a contour integral into infinitely many small circular contour integrals around every point, and most of them are zero, except a few that encircle singularities. Those are called residues, and add them up. This is basically how residue theorem works.
However, when I try to prove it by Green’s theorem, I face some difficulties. ($f=u+iv$)
$$oint_C f(z)dz=oint(u+iv)dx+i(u+iv)dy=iint_D fracpartial (u+iv)partial x-ifracpartial (u+iv)partial ydxdy$$
I don’t see how can this be transformed into a sum of residues/circular contour integrals around singularities.
Can you please show how?
Thanks in advance.
complex-analysis
asked Jul 25 at 3:47
Szeto
4,0931521
asked Jul 25 at 3:47
Szeto
4,0931521
asked Jul 25 at 3:47
Szeto
4,0931521
asked Jul 25 at 3:47
Szeto
4,0931521
4,0931521
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You are given a region $Omegasubsetmathbb C$ with boundary cycle $partialOmega$ and a function $f:>Omegatomathbb C$ which is analytic apart from finitely many isolated singularities at the points $a_jinOmega$ $(1leq jleq m)$. The trick now is to draw tiny discs $B_j$ of radius $epsilon>0$ centered at these points. Then we consider the punched domain $Omega':=Omegasetminusbigcup_j=1^m B_j$ with boundary cycle
$$partialOmega'=partialOmega-sum_j=1^m partial B_j .$$
Note the minus signs here! The function $f$ is analytic in $Omega'$; therefore we may apply Green's, resp., Cauchy's, theorem to $f$ and $Omega'$. It follows that $int_partialOmega'f(z)>dz=0$, so that
$$int_partialOmegaf(z)>dz=sum_j=1^mint_partial B_jf(z)>dz .$$
Here the LHS is independent of $epsilon$. It follows that
$$int_partialOmegaf(z)>dz=sum_j=1^mlim_epsilonto0int_partial B_jf(z)>dz=2pi i sum_j=1^mrm res(f, a_j) ,$$
by definition of the residue.
answered Jul 25 at 13:05
Christian Blatter
163k7107306
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You are given a region $Omegasubsetmathbb C$ with boundary cycle $partialOmega$ and a function $f:>Omegatomathbb C$ which is analytic apart from finitely many isolated singularities at the points $a_jinOmega$ $(1leq jleq m)$. The trick now is to draw tiny discs $B_j$ of radius $epsilon>0$ centered at these points. Then we consider the punched domain $Omega':=Omegasetminusbigcup_j=1^m B_j$ with boundary cycle
$$partialOmega'=partialOmega-sum_j=1^m partial B_j .$$
Note the minus signs here! The function $f$ is analytic in $Omega'$; therefore we may apply Green's, resp., Cauchy's, theorem to $f$ and $Omega'$. It follows that $int_partialOmega'f(z)>dz=0$, so that
$$int_partialOmegaf(z)>dz=sum_j=1^mint_partial B_jf(z)>dz .$$
Here the LHS is independent of $epsilon$. It follows that
$$int_partialOmegaf(z)>dz=sum_j=1^mlim_epsilonto0int_partial B_jf(z)>dz=2pi i sum_j=1^mrm res(f, a_j) ,$$
by definition of the residue.
answered Jul 25 at 13:05
Christian Blatter
163k7107306
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
add a comment |Â
up vote
3
down vote
accepted
You are given a region $Omegasubsetmathbb C$ with boundary cycle $partialOmega$ and a function $f:>Omegatomathbb C$ which is analytic apart from finitely many isolated singularities at the points $a_jinOmega$ $(1leq jleq m)$. The trick now is to draw tiny discs $B_j$ of radius $epsilon>0$ centered at these points. Then we consider the punched domain $Omega':=Omegasetminusbigcup_j=1^m B_j$ with boundary cycle
$$partialOmega'=partialOmega-sum_j=1^m partial B_j .$$
Note the minus signs here! The function $f$ is analytic in $Omega'$; therefore we may apply Green's, resp., Cauchy's, theorem to $f$ and $Omega'$. It follows that $int_partialOmega'f(z)>dz=0$, so that
$$int_partialOmegaf(z)>dz=sum_j=1^mint_partial B_jf(z)>dz .$$
Here the LHS is independent of $epsilon$. It follows that
$$int_partialOmegaf(z)>dz=sum_j=1^mlim_epsilonto0int_partial B_jf(z)>dz=2pi i sum_j=1^mrm res(f, a_j) ,$$
by definition of the residue.
answered Jul 25 at 13:05
Christian Blatter
163k7107306
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You are given a region $Omegasubsetmathbb C$ with boundary cycle $partialOmega$ and a function $f:>Omegatomathbb C$ which is analytic apart from finitely many isolated singularities at the points $a_jinOmega$ $(1leq jleq m)$. The trick now is to draw tiny discs $B_j$ of radius $epsilon>0$ centered at these points. Then we consider the punched domain $Omega':=Omegasetminusbigcup_j=1^m B_j$ with boundary cycle
$$partialOmega'=partialOmega-sum_j=1^m partial B_j .$$
Note the minus signs here! The function $f$ is analytic in $Omega'$; therefore we may apply Green's, resp., Cauchy's, theorem to $f$ and $Omega'$. It follows that $int_partialOmega'f(z)>dz=0$, so that
$$int_partialOmegaf(z)>dz=sum_j=1^mint_partial B_jf(z)>dz .$$
Here the LHS is independent of $epsilon$. It follows that
$$int_partialOmegaf(z)>dz=sum_j=1^mlim_epsilonto0int_partial B_jf(z)>dz=2pi i sum_j=1^mrm res(f, a_j) ,$$
by definition of the residue.
answered Jul 25 at 13:05
Christian Blatter
163k7107306
You are given a region $Omegasubsetmathbb C$ with boundary cycle $partialOmega$ and a function $f:>Omegatomathbb C$ which is analytic apart from finitely many isolated singularities at the points $a_jinOmega$ $(1leq jleq m)$. The trick now is to draw tiny discs $B_j$ of radius $epsilon>0$ centered at these points. Then we consider the punched domain $Omega':=Omegasetminusbigcup_j=1^m B_j$ with boundary cycle
$$partialOmega'=partialOmega-sum_j=1^m partial B_j .$$
Note the minus signs here! The function $f$ is analytic in $Omega'$; therefore we may apply Green's, resp., Cauchy's, theorem to $f$ and $Omega'$. It follows that $int_partialOmega'f(z)>dz=0$, so that
$$int_partialOmegaf(z)>dz=sum_j=1^mint_partial B_jf(z)>dz .$$
Here the LHS is independent of $epsilon$. It follows that
$$int_partialOmegaf(z)>dz=sum_j=1^mlim_epsilonto0int_partial B_jf(z)>dz=2pi i sum_j=1^mrm res(f, a_j) ,$$
by definition of the residue.
answered Jul 25 at 13:05
Christian Blatter
163k7107306
answered Jul 25 at 13:05
Christian Blatter
163k7107306
answered Jul 25 at 13:05
Christian Blatter
163k7107306
answered Jul 25 at 13:05
Christian Blatter
163k7107306
163k7107306
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
add a comment |Â
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
This is brilliant. Just one thing: how can I derive with details and rigor that $$int_partialOmega+partial B=int_partialOmega+int_partial B$$ to ‘convince’ a starter in complex analysis? (I know this is almost trivial to you.)
– Szeto
Jul 25 at 13:59
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Also, to my surprise, I cannot google a similar proof. Indeed, a formal proof of residue theorem is quite rare on the internet.
– Szeto
Jul 25 at 14:02
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
Already $partialOmega$ might consist of several disjoint closed curves. If you accept the notion of boundary cycle as defined in your community, as well as the notion of integral over a chain (formal sum of smooth curves) then there is no problem with $partialOmega'=partialOmega-sum_jpartial B_j$.
– Christian Blatter
Jul 25 at 14:10
add a comment |Â
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