How does the fact that every integral around a toy contour vanishes imply that the function is holomorphic?Demonstrating that a non-constant holomorphic function attains its maximum on the boundary of its region?Prove that for a holomorphic function $f$ on $mathbbC$, if $-f(z) = f(frac1z)$, then the residue of $f$ at 0 is 0.How to compute this contour integral with a modulus sign in the integrand,Is the integral of any even complex function equal to $0$ on any contour?A Contour IntegralDouble Integral Formula for Holomorphic Function on the Unit Disc (Complex Plain)When is the square root of a polynomial holomorphic? And how do you take the contour integral?Holomorphic function on a square that vanishes on a sideDoes the residue always exist for an isolated singularity?Suppose $f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with a limit point , then $f=0$ .
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How does the fact that every integral around a toy contour vanishes imply that the function is holomorphic?
Demonstrating that a non-constant holomorphic function attains its maximum on the boundary of its region?Prove that for a holomorphic function $f$ on $mathbbC$, if $-f(z) = f(frac1z)$, then the residue of $f$ at 0 is 0.How to compute this contour integral with a modulus sign in the integrand,Is the integral of any even complex function equal to $0$ on any contour?A Contour IntegralDouble Integral Formula for Holomorphic Function on the Unit Disc (Complex Plain)When is the square root of a polynomial holomorphic? And how do you take the contour integral?Holomorphic function on a square that vanishes on a sideDoes the residue always exist for an isolated singularity?Suppose $f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with a limit point , then $f=0$ .
$begingroup$
I know that if a function is holomorphic in the enclosed domain, then it follows that the integral around the contour vanishes. However, my question is rather, how does the other direction follow? If for every toy contour in some region D one has that the integral of f(z) vanishes, how does it follow that f is holomorphic in D?
My approach was, if the integral vanishes always, then there exists a holomorphic function F(z) such that F´(z)=f(z) for every z in D, but I dont get how does it follow that the derivative is also a holomorphic function?
Also, given z=x+iy and f(z)=max(0, x), which is not holomorphic on the unit disc since the derivative limit does not exist at 0, I am having some trouble computing the loop integral to actually see how it is nonzero.
I guess this is somewhat related to the Laurent expansion and the residue, which is the only non holomorphic term in the expansion but I am a litlle bit confused still.
Any help is appreciated, thank you
complex-analysis residue-calculus holomorphic-functions cauchy-integral-formula
New contributor
$endgroup$
add a comment |
$begingroup$
I know that if a function is holomorphic in the enclosed domain, then it follows that the integral around the contour vanishes. However, my question is rather, how does the other direction follow? If for every toy contour in some region D one has that the integral of f(z) vanishes, how does it follow that f is holomorphic in D?
My approach was, if the integral vanishes always, then there exists a holomorphic function F(z) such that F´(z)=f(z) for every z in D, but I dont get how does it follow that the derivative is also a holomorphic function?
Also, given z=x+iy and f(z)=max(0, x), which is not holomorphic on the unit disc since the derivative limit does not exist at 0, I am having some trouble computing the loop integral to actually see how it is nonzero.
I guess this is somewhat related to the Laurent expansion and the residue, which is the only non holomorphic term in the expansion but I am a litlle bit confused still.
Any help is appreciated, thank you
complex-analysis residue-calculus holomorphic-functions cauchy-integral-formula
New contributor
$endgroup$
$begingroup$
If a complex function is differentiable (holomorphic), then it is also analytic, i.e. it equals to its Taylor series, hence it is infinitely differentiable.
$endgroup$
– Berci
2 days ago
1
$begingroup$
Have you learned Morera's Theorem?
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Every book on Complex Analysis I have come across has Morera's Theorem. You are asking for a proof for a basic theorem.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
I saw the statement in another proof but it didnt explicitly say it was Moreras theorem. It just stated it and left me thinking about why. Thanks for your help
$endgroup$
– mmar
2 days ago
add a comment |
$begingroup$
I know that if a function is holomorphic in the enclosed domain, then it follows that the integral around the contour vanishes. However, my question is rather, how does the other direction follow? If for every toy contour in some region D one has that the integral of f(z) vanishes, how does it follow that f is holomorphic in D?
My approach was, if the integral vanishes always, then there exists a holomorphic function F(z) such that F´(z)=f(z) for every z in D, but I dont get how does it follow that the derivative is also a holomorphic function?
Also, given z=x+iy and f(z)=max(0, x), which is not holomorphic on the unit disc since the derivative limit does not exist at 0, I am having some trouble computing the loop integral to actually see how it is nonzero.
I guess this is somewhat related to the Laurent expansion and the residue, which is the only non holomorphic term in the expansion but I am a litlle bit confused still.
Any help is appreciated, thank you
complex-analysis residue-calculus holomorphic-functions cauchy-integral-formula
New contributor
$endgroup$
I know that if a function is holomorphic in the enclosed domain, then it follows that the integral around the contour vanishes. However, my question is rather, how does the other direction follow? If for every toy contour in some region D one has that the integral of f(z) vanishes, how does it follow that f is holomorphic in D?
My approach was, if the integral vanishes always, then there exists a holomorphic function F(z) such that F´(z)=f(z) for every z in D, but I dont get how does it follow that the derivative is also a holomorphic function?
Also, given z=x+iy and f(z)=max(0, x), which is not holomorphic on the unit disc since the derivative limit does not exist at 0, I am having some trouble computing the loop integral to actually see how it is nonzero.
I guess this is somewhat related to the Laurent expansion and the residue, which is the only non holomorphic term in the expansion but I am a litlle bit confused still.
Any help is appreciated, thank you
complex-analysis residue-calculus holomorphic-functions cauchy-integral-formula
complex-analysis residue-calculus holomorphic-functions cauchy-integral-formula
New contributor
New contributor
New contributor
asked 2 days ago
mmarmmar
255
255
New contributor
New contributor
$begingroup$
If a complex function is differentiable (holomorphic), then it is also analytic, i.e. it equals to its Taylor series, hence it is infinitely differentiable.
$endgroup$
– Berci
2 days ago
1
$begingroup$
Have you learned Morera's Theorem?
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Every book on Complex Analysis I have come across has Morera's Theorem. You are asking for a proof for a basic theorem.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
I saw the statement in another proof but it didnt explicitly say it was Moreras theorem. It just stated it and left me thinking about why. Thanks for your help
$endgroup$
– mmar
2 days ago
add a comment |
$begingroup$
If a complex function is differentiable (holomorphic), then it is also analytic, i.e. it equals to its Taylor series, hence it is infinitely differentiable.
$endgroup$
– Berci
2 days ago
1
$begingroup$
Have you learned Morera's Theorem?
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Every book on Complex Analysis I have come across has Morera's Theorem. You are asking for a proof for a basic theorem.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
I saw the statement in another proof but it didnt explicitly say it was Moreras theorem. It just stated it and left me thinking about why. Thanks for your help
$endgroup$
– mmar
2 days ago
$begingroup$
If a complex function is differentiable (holomorphic), then it is also analytic, i.e. it equals to its Taylor series, hence it is infinitely differentiable.
$endgroup$
– Berci
2 days ago
$begingroup$
If a complex function is differentiable (holomorphic), then it is also analytic, i.e. it equals to its Taylor series, hence it is infinitely differentiable.
$endgroup$
– Berci
2 days ago
1
1
$begingroup$
Have you learned Morera's Theorem?
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Have you learned Morera's Theorem?
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Every book on Complex Analysis I have come across has Morera's Theorem. You are asking for a proof for a basic theorem.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Every book on Complex Analysis I have come across has Morera's Theorem. You are asking for a proof for a basic theorem.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
I saw the statement in another proof but it didnt explicitly say it was Moreras theorem. It just stated it and left me thinking about why. Thanks for your help
$endgroup$
– mmar
2 days ago
$begingroup$
I saw the statement in another proof but it didnt explicitly say it was Moreras theorem. It just stated it and left me thinking about why. Thanks for your help
$endgroup$
– mmar
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
My approach was, if the integral vanishes always, then there exists a holomorphic function $F(z)$ such that $F'(z)=f(z)$ for every $z$ in $D$...
From there, we have the Cauchy integral formula:
$$F''(a) = frac22pi ioint_gammafracF(z)(z-a)^2,dz$$
where $gamma$ is some simple closed curve enclosing $a$. That formula includes that $F$ is twice differentiable (if we go through the proof), and thus that $f$ is differentiable.
Also, given $z=x+iy$ and $f(z)=max(0, x)$, which is not holomorphic on the unit disc...
I guess this is somewhat related to the Laurent expansion and the residue...
No. Those matters, the Laurent expansion and residues, are for dealing with functions which are holomorphic on a region except for some isolated points. Near those points, the function is invariably unbounded. That's not what's going on here.
In this case, we're looking at a function that's real-differentiable (except on the line $x=0$) but not complex-differentiable (on the right half where $x>0$). In order to evaluate the integral around a loop $gamma$ enclosing a region $R$ here, we use Green's theorem:
beginalign*oint_gammaf(z),dz &= int_a^b(g(X,Y)+ih(X,Y))cdot (X'(t)+iY'(t)),dt\
&= oint_gamma(g+ih)(X,Y),dx+(-h+ig)(X,Y),dy\
&= iint_R fracpartial(-h+ig)partial x-fracpartial(g+ih)partial y,dx,dyendalign*
For the function we're looking at, the real and imaginary parts $g$ and $h$ are $x$ and zero respectively in the right half $x>0$, for partial derivatives $fracpartial(-h+ig)partial x=i$ and $fracpartial(g+ih)partial y=0$ respectively. Integrate that, and we get $i$ times the area - or, at least, the portion of the area with $x>0$.
We could also calculate the line integrals explicitly with a nice enough region, such as a rectangle.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
My approach was, if the integral vanishes always, then there exists a holomorphic function $F(z)$ such that $F'(z)=f(z)$ for every $z$ in $D$...
From there, we have the Cauchy integral formula:
$$F''(a) = frac22pi ioint_gammafracF(z)(z-a)^2,dz$$
where $gamma$ is some simple closed curve enclosing $a$. That formula includes that $F$ is twice differentiable (if we go through the proof), and thus that $f$ is differentiable.
Also, given $z=x+iy$ and $f(z)=max(0, x)$, which is not holomorphic on the unit disc...
I guess this is somewhat related to the Laurent expansion and the residue...
No. Those matters, the Laurent expansion and residues, are for dealing with functions which are holomorphic on a region except for some isolated points. Near those points, the function is invariably unbounded. That's not what's going on here.
In this case, we're looking at a function that's real-differentiable (except on the line $x=0$) but not complex-differentiable (on the right half where $x>0$). In order to evaluate the integral around a loop $gamma$ enclosing a region $R$ here, we use Green's theorem:
beginalign*oint_gammaf(z),dz &= int_a^b(g(X,Y)+ih(X,Y))cdot (X'(t)+iY'(t)),dt\
&= oint_gamma(g+ih)(X,Y),dx+(-h+ig)(X,Y),dy\
&= iint_R fracpartial(-h+ig)partial x-fracpartial(g+ih)partial y,dx,dyendalign*
For the function we're looking at, the real and imaginary parts $g$ and $h$ are $x$ and zero respectively in the right half $x>0$, for partial derivatives $fracpartial(-h+ig)partial x=i$ and $fracpartial(g+ih)partial y=0$ respectively. Integrate that, and we get $i$ times the area - or, at least, the portion of the area with $x>0$.
We could also calculate the line integrals explicitly with a nice enough region, such as a rectangle.
$endgroup$
add a comment |
$begingroup$
My approach was, if the integral vanishes always, then there exists a holomorphic function $F(z)$ such that $F'(z)=f(z)$ for every $z$ in $D$...
From there, we have the Cauchy integral formula:
$$F''(a) = frac22pi ioint_gammafracF(z)(z-a)^2,dz$$
where $gamma$ is some simple closed curve enclosing $a$. That formula includes that $F$ is twice differentiable (if we go through the proof), and thus that $f$ is differentiable.
Also, given $z=x+iy$ and $f(z)=max(0, x)$, which is not holomorphic on the unit disc...
I guess this is somewhat related to the Laurent expansion and the residue...
No. Those matters, the Laurent expansion and residues, are for dealing with functions which are holomorphic on a region except for some isolated points. Near those points, the function is invariably unbounded. That's not what's going on here.
In this case, we're looking at a function that's real-differentiable (except on the line $x=0$) but not complex-differentiable (on the right half where $x>0$). In order to evaluate the integral around a loop $gamma$ enclosing a region $R$ here, we use Green's theorem:
beginalign*oint_gammaf(z),dz &= int_a^b(g(X,Y)+ih(X,Y))cdot (X'(t)+iY'(t)),dt\
&= oint_gamma(g+ih)(X,Y),dx+(-h+ig)(X,Y),dy\
&= iint_R fracpartial(-h+ig)partial x-fracpartial(g+ih)partial y,dx,dyendalign*
For the function we're looking at, the real and imaginary parts $g$ and $h$ are $x$ and zero respectively in the right half $x>0$, for partial derivatives $fracpartial(-h+ig)partial x=i$ and $fracpartial(g+ih)partial y=0$ respectively. Integrate that, and we get $i$ times the area - or, at least, the portion of the area with $x>0$.
We could also calculate the line integrals explicitly with a nice enough region, such as a rectangle.
$endgroup$
add a comment |
$begingroup$
My approach was, if the integral vanishes always, then there exists a holomorphic function $F(z)$ such that $F'(z)=f(z)$ for every $z$ in $D$...
From there, we have the Cauchy integral formula:
$$F''(a) = frac22pi ioint_gammafracF(z)(z-a)^2,dz$$
where $gamma$ is some simple closed curve enclosing $a$. That formula includes that $F$ is twice differentiable (if we go through the proof), and thus that $f$ is differentiable.
Also, given $z=x+iy$ and $f(z)=max(0, x)$, which is not holomorphic on the unit disc...
I guess this is somewhat related to the Laurent expansion and the residue...
No. Those matters, the Laurent expansion and residues, are for dealing with functions which are holomorphic on a region except for some isolated points. Near those points, the function is invariably unbounded. That's not what's going on here.
In this case, we're looking at a function that's real-differentiable (except on the line $x=0$) but not complex-differentiable (on the right half where $x>0$). In order to evaluate the integral around a loop $gamma$ enclosing a region $R$ here, we use Green's theorem:
beginalign*oint_gammaf(z),dz &= int_a^b(g(X,Y)+ih(X,Y))cdot (X'(t)+iY'(t)),dt\
&= oint_gamma(g+ih)(X,Y),dx+(-h+ig)(X,Y),dy\
&= iint_R fracpartial(-h+ig)partial x-fracpartial(g+ih)partial y,dx,dyendalign*
For the function we're looking at, the real and imaginary parts $g$ and $h$ are $x$ and zero respectively in the right half $x>0$, for partial derivatives $fracpartial(-h+ig)partial x=i$ and $fracpartial(g+ih)partial y=0$ respectively. Integrate that, and we get $i$ times the area - or, at least, the portion of the area with $x>0$.
We could also calculate the line integrals explicitly with a nice enough region, such as a rectangle.
$endgroup$
My approach was, if the integral vanishes always, then there exists a holomorphic function $F(z)$ such that $F'(z)=f(z)$ for every $z$ in $D$...
From there, we have the Cauchy integral formula:
$$F''(a) = frac22pi ioint_gammafracF(z)(z-a)^2,dz$$
where $gamma$ is some simple closed curve enclosing $a$. That formula includes that $F$ is twice differentiable (if we go through the proof), and thus that $f$ is differentiable.
Also, given $z=x+iy$ and $f(z)=max(0, x)$, which is not holomorphic on the unit disc...
I guess this is somewhat related to the Laurent expansion and the residue...
No. Those matters, the Laurent expansion and residues, are for dealing with functions which are holomorphic on a region except for some isolated points. Near those points, the function is invariably unbounded. That's not what's going on here.
In this case, we're looking at a function that's real-differentiable (except on the line $x=0$) but not complex-differentiable (on the right half where $x>0$). In order to evaluate the integral around a loop $gamma$ enclosing a region $R$ here, we use Green's theorem:
beginalign*oint_gammaf(z),dz &= int_a^b(g(X,Y)+ih(X,Y))cdot (X'(t)+iY'(t)),dt\
&= oint_gamma(g+ih)(X,Y),dx+(-h+ig)(X,Y),dy\
&= iint_R fracpartial(-h+ig)partial x-fracpartial(g+ih)partial y,dx,dyendalign*
For the function we're looking at, the real and imaginary parts $g$ and $h$ are $x$ and zero respectively in the right half $x>0$, for partial derivatives $fracpartial(-h+ig)partial x=i$ and $fracpartial(g+ih)partial y=0$ respectively. Integrate that, and we get $i$ times the area - or, at least, the portion of the area with $x>0$.
We could also calculate the line integrals explicitly with a nice enough region, such as a rectangle.
answered 2 days ago
jmerryjmerry
13.3k1628
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add a comment |
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mmar is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
If a complex function is differentiable (holomorphic), then it is also analytic, i.e. it equals to its Taylor series, hence it is infinitely differentiable.
$endgroup$
– Berci
2 days ago
1
$begingroup$
Have you learned Morera's Theorem?
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Every book on Complex Analysis I have come across has Morera's Theorem. You are asking for a proof for a basic theorem.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
I saw the statement in another proof but it didnt explicitly say it was Moreras theorem. It just stated it and left me thinking about why. Thanks for your help
$endgroup$
– mmar
2 days ago