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How to understand the notation $ fracpartial fpartial overlinez $
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Suppose $ f(z):mathbbC^1tomathbbC^1 $, write $ fracpartial fpartial overlinez=frac12(fracpartial fpartial x+ifracpartial fpartial y) $. We know that by Cauchy-Riemann equation, the differential property is equivalent to $ fracpartial fpartial overlinez=0 $. But can we take the derivative directly regarding to $ overlinez $ to determine whether the function is differential or not? If we can't do this, then what's the reason to introduce this symbol? And how to determine a function is differentiale without directly expanding the real and imaginary part out and check the Cauchy-Riemann equation?
complex-analysis analysis
asked Jul 19 at 11:31
Yuchen
751114
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up vote
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Suppose $ f(z):mathbbC^1tomathbbC^1 $, write $ fracpartial fpartial overlinez=frac12(fracpartial fpartial x+ifracpartial fpartial y) $. We know that by Cauchy-Riemann equation, the differential property is equivalent to $ fracpartial fpartial overlinez=0 $. But can we take the derivative directly regarding to $ overlinez $ to determine whether the function is differential or not? If we can't do this, then what's the reason to introduce this symbol? And how to determine a function is differentiale without directly expanding the real and imaginary part out and check the Cauchy-Riemann equation?
complex-analysis analysis
asked Jul 19 at 11:31
Yuchen
751114
1
Note that $fracpartial fpartial z$ and $fracpartial fpartial bar z$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $bar z$, but rather the result of applying to $f$ the following operators:$$fracpartialpartial z=frac12left(fracpartialpartial x+frac1ifracpartialpartial yright),quadfracpartialpartial bar z=frac12left(fracpartialpartial x-frac1ifracpartialpartial yright).$$ -- Berenstein C. A., Complex Variables, An Introduction. $nwarrow$ relevant book snippet
– Frenzy Li
Jul 19 at 11:39
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up vote
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down vote
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up vote
3
down vote
favorite
Suppose $ f(z):mathbbC^1tomathbbC^1 $, write $ fracpartial fpartial overlinez=frac12(fracpartial fpartial x+ifracpartial fpartial y) $. We know that by Cauchy-Riemann equation, the differential property is equivalent to $ fracpartial fpartial overlinez=0 $. But can we take the derivative directly regarding to $ overlinez $ to determine whether the function is differential or not? If we can't do this, then what's the reason to introduce this symbol? And how to determine a function is differentiale without directly expanding the real and imaginary part out and check the Cauchy-Riemann equation?
complex-analysis analysis
asked Jul 19 at 11:31
Yuchen
751114
Suppose $ f(z):mathbbC^1tomathbbC^1 $, write $ fracpartial fpartial overlinez=frac12(fracpartial fpartial x+ifracpartial fpartial y) $. We know that by Cauchy-Riemann equation, the differential property is equivalent to $ fracpartial fpartial overlinez=0 $. But can we take the derivative directly regarding to $ overlinez $ to determine whether the function is differential or not? If we can't do this, then what's the reason to introduce this symbol? And how to determine a function is differentiale without directly expanding the real and imaginary part out and check the Cauchy-Riemann equation?
complex-analysis analysis
asked Jul 19 at 11:31
Yuchen
751114
edited Jul 19 at 12:43
asked Jul 19 at 11:31
Yuchen
751114
asked Jul 19 at 11:31
Yuchen
751114
asked Jul 19 at 11:31
Yuchen
751114
751114
1
Note that $fracpartial fpartial z$ and $fracpartial fpartial bar z$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $bar z$, but rather the result of applying to $f$ the following operators:$$fracpartialpartial z=frac12left(fracpartialpartial x+frac1ifracpartialpartial yright),quadfracpartialpartial bar z=frac12left(fracpartialpartial x-frac1ifracpartialpartial yright).$$ -- Berenstein C. A., Complex Variables, An Introduction. $nwarrow$ relevant book snippet
– Frenzy Li
Jul 19 at 11:39
add a comment |Â
1
Note that $fracpartial fpartial z$ and $fracpartial fpartial bar z$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $bar z$, but rather the result of applying to $f$ the following operators:$$fracpartialpartial z=frac12left(fracpartialpartial x+frac1ifracpartialpartial yright),quadfracpartialpartial bar z=frac12left(fracpartialpartial x-frac1ifracpartialpartial yright).$$ -- Berenstein C. A., Complex Variables, An Introduction. $nwarrow$ relevant book snippet
– Frenzy Li
Jul 19 at 11:39
1
1
Note that $fracpartial fpartial z$ and $fracpartial fpartial bar z$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $bar z$, but rather the result of applying to $f$ the following operators:$$fracpartialpartial z=frac12left(fracpartialpartial x+frac1ifracpartialpartial yright),quadfracpartialpartial bar z=frac12left(fracpartialpartial x-frac1ifracpartialpartial yright).$$ -- Berenstein C. A., Complex Variables, An Introduction. $nwarrow$ relevant book snippet
– Frenzy Li
Jul 19 at 11:39
Note that $fracpartial fpartial z$ and $fracpartial fpartial bar z$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $bar z$, but rather the result of applying to $f$ the following operators:$$fracpartialpartial z=frac12left(fracpartialpartial x+frac1ifracpartialpartial yright),quadfracpartialpartial bar z=frac12left(fracpartialpartial x-frac1ifracpartialpartial yright).$$ -- Berenstein C. A., Complex Variables, An Introduction. $nwarrow$ relevant book snippet
– Frenzy Li
Jul 19 at 11:39
add a comment |Â
3 Answers
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There is the complex vector space $V$ of real-linear complex valued functions
$$f:quadmathbb R^2tomathbb C,qquadbf z=(x,y)to f(bf z) .$$
Such functions satisfy
$$f(bf z+bf w)=f(bf z)+f(bf w),qquad f(lambdabf z)=lambda,f(bf z)quad(lambdainmathbb R) .$$
It is easy to see that $rm dim_mathbb C(V)=2$, and that the two functions $$x:quad (x,y)mapsto x, qquad y:quad (x,y)mapsto y$$ (here we see some abuse of notation) form a basis of $V$, meaning that each $fin V$ has a unique representation $$f(x,y)=c_1 x+c_2 y,qquad c_1,c_2inmathbb C .$$
The partial differential operators $partialoverpartial x$ and $partialoverpartial y$ then form the corresponding dual basis of $V^*$, meaning that $c_1=partialoverpartial xf$, $>c_2=partialoverpartial yf$.
Now the two functions $z:=x+iyin V$ and $bar z:=x-iyin V$ form a basis of $V$ as well, and the Wirtinger derivatives introduced in the question form the corresponding dual basis of $V^*$, meaning that for $f=c_1z+c_2bar zin V$ one has $c_1=partialoverpartial zf$, $>c_2=partialoverpartialbar zf$.
answered Jul 19 at 13:40
Christian Blatter
163k7107306
1
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
add a comment |Â
up vote
6
down vote
Is somewhat of an abuse of notation, but the idea is that if $f:mathbb Ctomathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write
$$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$
where $F(h)$ is a linear transformation of $mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written
$$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$
for some (uniquely determined) complex constants $c_1$ and $c_2$.
Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = overline z - overlinez_0 $, there is a certain justice in calling them $partial f/partial z$ and $partial f/partial bar z$.
But don't take this for more than it is; the notation depends on pretending that $z$ and $bar z$ can vary independently of each other, which is not really the case.
answered Jul 19 at 11:47
Henning Makholm
226k16291519
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
add a comment |Â
up vote
1
down vote
The differential operator
$$fracpartialpartialoverline z$$
has the property that
$$fracpartialpartialoverline zz=0qquadtextrmand
qquadfracpartialpartialoverline zoverline z=1.$$
Likewise the differential operator
$$fracpartialpartial z=frac12left(
fracpartialpartial x-ifracpartialpartial yright)$$
has the property that
$$fracpartialpartial zz=1qquadtextrmand
qquadfracpartialpartial zoverline z=0.$$
As $partial/partial overline z$ is an elliptic operator, any distribution it annihilates is a smooth function satisfying Cauchy-Riemann and so is a holomorphic function.
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is the complex vector space $V$ of real-linear complex valued functions
$$f:quadmathbb R^2tomathbb C,qquadbf z=(x,y)to f(bf z) .$$
Such functions satisfy
$$f(bf z+bf w)=f(bf z)+f(bf w),qquad f(lambdabf z)=lambda,f(bf z)quad(lambdainmathbb R) .$$
It is easy to see that $rm dim_mathbb C(V)=2$, and that the two functions $$x:quad (x,y)mapsto x, qquad y:quad (x,y)mapsto y$$ (here we see some abuse of notation) form a basis of $V$, meaning that each $fin V$ has a unique representation $$f(x,y)=c_1 x+c_2 y,qquad c_1,c_2inmathbb C .$$
The partial differential operators $partialoverpartial x$ and $partialoverpartial y$ then form the corresponding dual basis of $V^*$, meaning that $c_1=partialoverpartial xf$, $>c_2=partialoverpartial yf$.
Now the two functions $z:=x+iyin V$ and $bar z:=x-iyin V$ form a basis of $V$ as well, and the Wirtinger derivatives introduced in the question form the corresponding dual basis of $V^*$, meaning that for $f=c_1z+c_2bar zin V$ one has $c_1=partialoverpartial zf$, $>c_2=partialoverpartialbar zf$.
answered Jul 19 at 13:40
Christian Blatter
163k7107306
1
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
add a comment |Â
up vote
1
down vote
accepted
There is the complex vector space $V$ of real-linear complex valued functions
$$f:quadmathbb R^2tomathbb C,qquadbf z=(x,y)to f(bf z) .$$
Such functions satisfy
$$f(bf z+bf w)=f(bf z)+f(bf w),qquad f(lambdabf z)=lambda,f(bf z)quad(lambdainmathbb R) .$$
It is easy to see that $rm dim_mathbb C(V)=2$, and that the two functions $$x:quad (x,y)mapsto x, qquad y:quad (x,y)mapsto y$$ (here we see some abuse of notation) form a basis of $V$, meaning that each $fin V$ has a unique representation $$f(x,y)=c_1 x+c_2 y,qquad c_1,c_2inmathbb C .$$
The partial differential operators $partialoverpartial x$ and $partialoverpartial y$ then form the corresponding dual basis of $V^*$, meaning that $c_1=partialoverpartial xf$, $>c_2=partialoverpartial yf$.
Now the two functions $z:=x+iyin V$ and $bar z:=x-iyin V$ form a basis of $V$ as well, and the Wirtinger derivatives introduced in the question form the corresponding dual basis of $V^*$, meaning that for $f=c_1z+c_2bar zin V$ one has $c_1=partialoverpartial zf$, $>c_2=partialoverpartialbar zf$.
answered Jul 19 at 13:40
Christian Blatter
163k7107306
1
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is the complex vector space $V$ of real-linear complex valued functions
$$f:quadmathbb R^2tomathbb C,qquadbf z=(x,y)to f(bf z) .$$
Such functions satisfy
$$f(bf z+bf w)=f(bf z)+f(bf w),qquad f(lambdabf z)=lambda,f(bf z)quad(lambdainmathbb R) .$$
It is easy to see that $rm dim_mathbb C(V)=2$, and that the two functions $$x:quad (x,y)mapsto x, qquad y:quad (x,y)mapsto y$$ (here we see some abuse of notation) form a basis of $V$, meaning that each $fin V$ has a unique representation $$f(x,y)=c_1 x+c_2 y,qquad c_1,c_2inmathbb C .$$
The partial differential operators $partialoverpartial x$ and $partialoverpartial y$ then form the corresponding dual basis of $V^*$, meaning that $c_1=partialoverpartial xf$, $>c_2=partialoverpartial yf$.
Now the two functions $z:=x+iyin V$ and $bar z:=x-iyin V$ form a basis of $V$ as well, and the Wirtinger derivatives introduced in the question form the corresponding dual basis of $V^*$, meaning that for $f=c_1z+c_2bar zin V$ one has $c_1=partialoverpartial zf$, $>c_2=partialoverpartialbar zf$.
answered Jul 19 at 13:40
Christian Blatter
163k7107306
There is the complex vector space $V$ of real-linear complex valued functions
$$f:quadmathbb R^2tomathbb C,qquadbf z=(x,y)to f(bf z) .$$
Such functions satisfy
$$f(bf z+bf w)=f(bf z)+f(bf w),qquad f(lambdabf z)=lambda,f(bf z)quad(lambdainmathbb R) .$$
It is easy to see that $rm dim_mathbb C(V)=2$, and that the two functions $$x:quad (x,y)mapsto x, qquad y:quad (x,y)mapsto y$$ (here we see some abuse of notation) form a basis of $V$, meaning that each $fin V$ has a unique representation $$f(x,y)=c_1 x+c_2 y,qquad c_1,c_2inmathbb C .$$
The partial differential operators $partialoverpartial x$ and $partialoverpartial y$ then form the corresponding dual basis of $V^*$, meaning that $c_1=partialoverpartial xf$, $>c_2=partialoverpartial yf$.
Now the two functions $z:=x+iyin V$ and $bar z:=x-iyin V$ form a basis of $V$ as well, and the Wirtinger derivatives introduced in the question form the corresponding dual basis of $V^*$, meaning that for $f=c_1z+c_2bar zin V$ one has $c_1=partialoverpartial zf$, $>c_2=partialoverpartialbar zf$.
answered Jul 19 at 13:40
Christian Blatter
163k7107306
answered Jul 19 at 13:40
Christian Blatter
163k7107306
answered Jul 19 at 13:40
Christian Blatter
163k7107306
answered Jul 19 at 13:40
Christian Blatter
163k7107306
163k7107306
1
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
add a comment |Â
1
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
1
1
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra.
– Christian Blatter
Jul 19 at 13:54
add a comment |Â
up vote
6
down vote
Is somewhat of an abuse of notation, but the idea is that if $f:mathbb Ctomathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write
$$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$
where $F(h)$ is a linear transformation of $mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written
$$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$
for some (uniquely determined) complex constants $c_1$ and $c_2$.
Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = overline z - overlinez_0 $, there is a certain justice in calling them $partial f/partial z$ and $partial f/partial bar z$.
But don't take this for more than it is; the notation depends on pretending that $z$ and $bar z$ can vary independently of each other, which is not really the case.
answered Jul 19 at 11:47
Henning Makholm
226k16291519
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
add a comment |Â
up vote
6
down vote
Is somewhat of an abuse of notation, but the idea is that if $f:mathbb Ctomathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write
$$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$
where $F(h)$ is a linear transformation of $mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written
$$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$
for some (uniquely determined) complex constants $c_1$ and $c_2$.
Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = overline z - overlinez_0 $, there is a certain justice in calling them $partial f/partial z$ and $partial f/partial bar z$.
But don't take this for more than it is; the notation depends on pretending that $z$ and $bar z$ can vary independently of each other, which is not really the case.
answered Jul 19 at 11:47
Henning Makholm
226k16291519
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Is somewhat of an abuse of notation, but the idea is that if $f:mathbb Ctomathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write
$$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$
where $F(h)$ is a linear transformation of $mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written
$$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$
for some (uniquely determined) complex constants $c_1$ and $c_2$.
Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = overline z - overlinez_0 $, there is a certain justice in calling them $partial f/partial z$ and $partial f/partial bar z$.
But don't take this for more than it is; the notation depends on pretending that $z$ and $bar z$ can vary independently of each other, which is not really the case.
answered Jul 19 at 11:47
Henning Makholm
226k16291519
Is somewhat of an abuse of notation, but the idea is that if $f:mathbb Ctomathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write
$$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$
where $F(h)$ is a linear transformation of $mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written
$$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$
for some (uniquely determined) complex constants $c_1$ and $c_2$.
Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = overline z - overlinez_0 $, there is a certain justice in calling them $partial f/partial z$ and $partial f/partial bar z$.
But don't take this for more than it is; the notation depends on pretending that $z$ and $bar z$ can vary independently of each other, which is not really the case.
answered Jul 19 at 11:47
Henning Makholm
226k16291519
answered Jul 19 at 11:47
Henning Makholm
226k16291519
answered Jul 19 at 11:47
Henning Makholm
226k16291519
answered Jul 19 at 11:47
Henning Makholm
226k16291519
226k16291519
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
add a comment |Â
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
Somewhat of an abuse of notation?!
– Umberto P.
Jul 19 at 11:53
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity.
– Henning Makholm
Jul 19 at 11:57
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
@UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $mathbb C$, and then $z$ and $overlinez$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level.
– paul garrett
Jul 19 at 12:55
add a comment |Â
up vote
1
down vote
The differential operator
$$fracpartialpartialoverline z$$
has the property that
$$fracpartialpartialoverline zz=0qquadtextrmand
qquadfracpartialpartialoverline zoverline z=1.$$
Likewise the differential operator
$$fracpartialpartial z=frac12left(
fracpartialpartial x-ifracpartialpartial yright)$$
has the property that
$$fracpartialpartial zz=1qquadtextrmand
qquadfracpartialpartial zoverline z=0.$$
As $partial/partial overline z$ is an elliptic operator, any distribution it annihilates is a smooth function satisfying Cauchy-Riemann and so is a holomorphic function.
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
add a comment |Â
up vote
1
down vote
The differential operator
$$fracpartialpartialoverline z$$
has the property that
$$fracpartialpartialoverline zz=0qquadtextrmand
qquadfracpartialpartialoverline zoverline z=1.$$
Likewise the differential operator
$$fracpartialpartial z=frac12left(
fracpartialpartial x-ifracpartialpartial yright)$$
has the property that
$$fracpartialpartial zz=1qquadtextrmand
qquadfracpartialpartial zoverline z=0.$$
As $partial/partial overline z$ is an elliptic operator, any distribution it annihilates is a smooth function satisfying Cauchy-Riemann and so is a holomorphic function.
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The differential operator
$$fracpartialpartialoverline z$$
has the property that
$$fracpartialpartialoverline zz=0qquadtextrmand
qquadfracpartialpartialoverline zoverline z=1.$$
Likewise the differential operator
$$fracpartialpartial z=frac12left(
fracpartialpartial x-ifracpartialpartial yright)$$
has the property that
$$fracpartialpartial zz=1qquadtextrmand
qquadfracpartialpartial zoverline z=0.$$
As $partial/partial overline z$ is an elliptic operator, any distribution it annihilates is a smooth function satisfying Cauchy-Riemann and so is a holomorphic function.
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
The differential operator
$$fracpartialpartialoverline z$$
has the property that
$$fracpartialpartialoverline zz=0qquadtextrmand
qquadfracpartialpartialoverline zoverline z=1.$$
Likewise the differential operator
$$fracpartialpartial z=frac12left(
fracpartialpartial x-ifracpartialpartial yright)$$
has the property that
$$fracpartialpartial zz=1qquadtextrmand
qquadfracpartialpartial zoverline z=0.$$
As $partial/partial overline z$ is an elliptic operator, any distribution it annihilates is a smooth function satisfying Cauchy-Riemann and so is a holomorphic function.
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:36
Lord Shark the Unknown
85.5k951112
85.5k951112
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1
Note that $fracpartial fpartial z$ and $fracpartial fpartial bar z$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $bar z$, but rather the result of applying to $f$ the following operators:$$fracpartialpartial z=frac12left(fracpartialpartial x+frac1ifracpartialpartial yright),quadfracpartialpartial bar z=frac12left(fracpartialpartial x-frac1ifracpartialpartial yright).$$ -- Berenstein C. A., Complex Variables, An Introduction. $nwarrow$ relevant book snippet
– Frenzy Li
Jul 19 at 11:39