Integral over complex conjugate domainDifferences between Cauchy integral theorem and fundamental theorem for integral calculus over a cycleComputing complex integralSimple complex line integral over a rectangleCauchy integral theorem: Is analyticity enough? Do we need a simply connected domain?Proving a version of Cauchy's theoremComplex Calculus question relating to Moreras theorem.Morera Theorem And Cauchy's Integral TheoremIntegral of a complex function over contourIntegral of a complex function over contour (continued)When would I use Cauchy's Integral Formula over Residue
Hotkey (or other quick way) to insert a keyframe for only one component of a vector-valued property?
Why the color red for the Republican Party
weren't playing vs didn't play
Definition of Statistic
Are tamper resistant receptacles really safer?
An alternative proof of an application of Hahn-Banach
how to copy/paste a formula in Excel absolutely?
How can I ensure my trip to the UK will not have to be cancelled because of Brexit?
Does the nature of the Apocalypse in The Umbrella Academy change from the first to the last episode?
Virginia employer terminated employee and wants signing bonus returned
What are actual Tesla M60 models used by AWS?
How can The Temple of Elementary Evil reliably protect itself against kinetic bombardment?
PTIJ: Should I kill my computer after installing software?
'The literal of type int is out of range' con número enteros pequeños (2 dígitos)
Does this video of collapsing warehouse shelves show a real incident?
Intuition behind counterexample of Euler's sum of powers conjecture
How is the wildcard * interpreted as a command?
Contract Factories
Is it possible to avoid unpacking when merging Association?
What's the "normal" opposite of flautando?
Was Luke Skywalker the leader of the Rebel forces on Hoth?
Do f-stop and exposure time perfectly cancel?
How to draw cubes in a 3 dimensional plane
Are all players supposed to be able to see each others' character sheets?
Integral over complex conjugate domain
Differences between Cauchy integral theorem and fundamental theorem for integral calculus over a cycleComputing complex integralSimple complex line integral over a rectangleCauchy integral theorem: Is analyticity enough? Do we need a simply connected domain?Proving a version of Cauchy's theoremComplex Calculus question relating to Moreras theorem.Morera Theorem And Cauchy's Integral TheoremIntegral of a complex function over contourIntegral of a complex function over contour (continued)When would I use Cauchy's Integral Formula over Residue
$begingroup$
So I know that $e^ikz,~zinmathbbC$ is analytic, so without any poles we have
$$int_gamma e^ikzdz=0$$
for any path $gamma$. But what if the integration is taken over the conjugate domain, like
$$int_gamma e^ikzdbarz?$$
Is it simple to come up with a residue theorem for this, or does one have to re-engineer Cauchy's theorem for every specific case?
complex-analysis
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
$endgroup$
add a comment |
$begingroup$
So I know that $e^ikz,~zinmathbbC$ is analytic, so without any poles we have
$$int_gamma e^ikzdz=0$$
for any path $gamma$. But what if the integration is taken over the conjugate domain, like
$$int_gamma e^ikzdbarz?$$
Is it simple to come up with a residue theorem for this, or does one have to re-engineer Cauchy's theorem for every specific case?
complex-analysis
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
$endgroup$
add a comment |
$begingroup$
So I know that $e^ikz,~zinmathbbC$ is analytic, so without any poles we have
$$int_gamma e^ikzdz=0$$
for any path $gamma$. But what if the integration is taken over the conjugate domain, like
$$int_gamma e^ikzdbarz?$$
Is it simple to come up with a residue theorem for this, or does one have to re-engineer Cauchy's theorem for every specific case?
complex-analysis
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
$endgroup$
So I know that $e^ikz,~zinmathbbC$ is analytic, so without any poles we have
$$int_gamma e^ikzdz=0$$
for any path $gamma$. But what if the integration is taken over the conjugate domain, like
$$int_gamma e^ikzdbarz?$$
Is it simple to come up with a residue theorem for this, or does one have to re-engineer Cauchy's theorem for every specific case?
complex-analysis
complex-analysis
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
asked Feb 8 at 21:17
levitopherlevitopher
1,1161226
1,1161226
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Applying the conjugate there is essentially equivalent to applying it to the function; if $I(gamma) = int_gammaf(z),doverlinez$, then $overlineI(gamma) = int_gammaoverlinef(z),dz$.
And the problem with that? It's strongly path-dependent. Unlike holomorphic functions, conjugate-holomorphic functions aren't exact forms. Let $f(x+iy)=g(x,y)+ih(x,y)$ and parametrize the closed loop $gamma$ enclosing a region $R$ by $Z(t)=X(t)+iY(t)$ for $t$ from $a$ to $b$. We get
beginalign*I(gamma) &= int_a^b left(g(X(t),Y(t))+ih(X(t),Y(t)right)cdot (X'(t)-iY'(t)),dt\
&= int_a^b g(X,Y)X'(t) + h(X,Y)Y'(t) + ih(X,Y)X'(t)-ig(X,Y)Y'(t),dt\
&= int_gamma(g+ih)(X,Y),dx + (h-ig)(X,Y),dy\
I(gamma) &= iint_R fracpartial (h-ig)partial x - fracpartial (g+ih)partial y,dx,dy\
&= iint_R -if'(x+iy) - if'(x+iy),dx,dyendalign*
(We abbreviate $g(X(t),Y(t))$ by $g(X,Y)$. These are all real double integrals and line integrals, although we allow complex-valued functions to simplify things a bit. The step from a line integral to an area integral was an application of Green's theorem.)
So then, the integral $int_gamma f(z),doverlinez$ is the area integral of $-2if'$ over the region enclosed. It's not something that we can focus on a few points to handle - we have to deal with the whole region. And that's for a function that doesn't have any singularities in there.
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3105612%2fintegral-over-complex-conjugate-domain%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Applying the conjugate there is essentially equivalent to applying it to the function; if $I(gamma) = int_gammaf(z),doverlinez$, then $overlineI(gamma) = int_gammaoverlinef(z),dz$.
And the problem with that? It's strongly path-dependent. Unlike holomorphic functions, conjugate-holomorphic functions aren't exact forms. Let $f(x+iy)=g(x,y)+ih(x,y)$ and parametrize the closed loop $gamma$ enclosing a region $R$ by $Z(t)=X(t)+iY(t)$ for $t$ from $a$ to $b$. We get
beginalign*I(gamma) &= int_a^b left(g(X(t),Y(t))+ih(X(t),Y(t)right)cdot (X'(t)-iY'(t)),dt\
&= int_a^b g(X,Y)X'(t) + h(X,Y)Y'(t) + ih(X,Y)X'(t)-ig(X,Y)Y'(t),dt\
&= int_gamma(g+ih)(X,Y),dx + (h-ig)(X,Y),dy\
I(gamma) &= iint_R fracpartial (h-ig)partial x - fracpartial (g+ih)partial y,dx,dy\
&= iint_R -if'(x+iy) - if'(x+iy),dx,dyendalign*
(We abbreviate $g(X(t),Y(t))$ by $g(X,Y)$. These are all real double integrals and line integrals, although we allow complex-valued functions to simplify things a bit. The step from a line integral to an area integral was an application of Green's theorem.)
So then, the integral $int_gamma f(z),doverlinez$ is the area integral of $-2if'$ over the region enclosed. It's not something that we can focus on a few points to handle - we have to deal with the whole region. And that's for a function that doesn't have any singularities in there.
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
$endgroup$
add a comment |
$begingroup$
Applying the conjugate there is essentially equivalent to applying it to the function; if $I(gamma) = int_gammaf(z),doverlinez$, then $overlineI(gamma) = int_gammaoverlinef(z),dz$.
And the problem with that? It's strongly path-dependent. Unlike holomorphic functions, conjugate-holomorphic functions aren't exact forms. Let $f(x+iy)=g(x,y)+ih(x,y)$ and parametrize the closed loop $gamma$ enclosing a region $R$ by $Z(t)=X(t)+iY(t)$ for $t$ from $a$ to $b$. We get
beginalign*I(gamma) &= int_a^b left(g(X(t),Y(t))+ih(X(t),Y(t)right)cdot (X'(t)-iY'(t)),dt\
&= int_a^b g(X,Y)X'(t) + h(X,Y)Y'(t) + ih(X,Y)X'(t)-ig(X,Y)Y'(t),dt\
&= int_gamma(g+ih)(X,Y),dx + (h-ig)(X,Y),dy\
I(gamma) &= iint_R fracpartial (h-ig)partial x - fracpartial (g+ih)partial y,dx,dy\
&= iint_R -if'(x+iy) - if'(x+iy),dx,dyendalign*
(We abbreviate $g(X(t),Y(t))$ by $g(X,Y)$. These are all real double integrals and line integrals, although we allow complex-valued functions to simplify things a bit. The step from a line integral to an area integral was an application of Green's theorem.)
So then, the integral $int_gamma f(z),doverlinez$ is the area integral of $-2if'$ over the region enclosed. It's not something that we can focus on a few points to handle - we have to deal with the whole region. And that's for a function that doesn't have any singularities in there.
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
$endgroup$
add a comment |
$begingroup$
Applying the conjugate there is essentially equivalent to applying it to the function; if $I(gamma) = int_gammaf(z),doverlinez$, then $overlineI(gamma) = int_gammaoverlinef(z),dz$.
And the problem with that? It's strongly path-dependent. Unlike holomorphic functions, conjugate-holomorphic functions aren't exact forms. Let $f(x+iy)=g(x,y)+ih(x,y)$ and parametrize the closed loop $gamma$ enclosing a region $R$ by $Z(t)=X(t)+iY(t)$ for $t$ from $a$ to $b$. We get
beginalign*I(gamma) &= int_a^b left(g(X(t),Y(t))+ih(X(t),Y(t)right)cdot (X'(t)-iY'(t)),dt\
&= int_a^b g(X,Y)X'(t) + h(X,Y)Y'(t) + ih(X,Y)X'(t)-ig(X,Y)Y'(t),dt\
&= int_gamma(g+ih)(X,Y),dx + (h-ig)(X,Y),dy\
I(gamma) &= iint_R fracpartial (h-ig)partial x - fracpartial (g+ih)partial y,dx,dy\
&= iint_R -if'(x+iy) - if'(x+iy),dx,dyendalign*
(We abbreviate $g(X(t),Y(t))$ by $g(X,Y)$. These are all real double integrals and line integrals, although we allow complex-valued functions to simplify things a bit. The step from a line integral to an area integral was an application of Green's theorem.)
So then, the integral $int_gamma f(z),doverlinez$ is the area integral of $-2if'$ over the region enclosed. It's not something that we can focus on a few points to handle - we have to deal with the whole region. And that's for a function that doesn't have any singularities in there.
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
$endgroup$
Applying the conjugate there is essentially equivalent to applying it to the function; if $I(gamma) = int_gammaf(z),doverlinez$, then $overlineI(gamma) = int_gammaoverlinef(z),dz$.
And the problem with that? It's strongly path-dependent. Unlike holomorphic functions, conjugate-holomorphic functions aren't exact forms. Let $f(x+iy)=g(x,y)+ih(x,y)$ and parametrize the closed loop $gamma$ enclosing a region $R$ by $Z(t)=X(t)+iY(t)$ for $t$ from $a$ to $b$. We get
beginalign*I(gamma) &= int_a^b left(g(X(t),Y(t))+ih(X(t),Y(t)right)cdot (X'(t)-iY'(t)),dt\
&= int_a^b g(X,Y)X'(t) + h(X,Y)Y'(t) + ih(X,Y)X'(t)-ig(X,Y)Y'(t),dt\
&= int_gamma(g+ih)(X,Y),dx + (h-ig)(X,Y),dy\
I(gamma) &= iint_R fracpartial (h-ig)partial x - fracpartial (g+ih)partial y,dx,dy\
&= iint_R -if'(x+iy) - if'(x+iy),dx,dyendalign*
(We abbreviate $g(X(t),Y(t))$ by $g(X,Y)$. These are all real double integrals and line integrals, although we allow complex-valued functions to simplify things a bit. The step from a line integral to an area integral was an application of Green's theorem.)
So then, the integral $int_gamma f(z),doverlinez$ is the area integral of $-2if'$ over the region enclosed. It's not something that we can focus on a few points to handle - we have to deal with the whole region. And that's for a function that doesn't have any singularities in there.
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
edited 2 days ago
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
answered Feb 8 at 21:53
jmerryjmerry
13.4k1628
13.4k1628
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3105612%2fintegral-over-complex-conjugate-domain%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
