Differentiating the complex scalar field

[Fredrik]

You should realize that the same word may have different meanings in different contexts, being generalized by mathematicians if they can give it a more general interpretation that still fits the formal rules.

Unless of course that word happens to be "observable", right?

Edit: You're obviously going to counter by saying that you're talking about a generalization while I was talking about a restriction. This means that you aren't contradicting yourself, but I still find it funny that you're so willing to embrace a redefinition of the term "independent" that assigns it to a pair of variables that have "I depend on that guy" tatooed on their foreheads, and at the same time find a restriction of the term "observable" so appalling.



I think I have a pretty good idea about how this Wirtinger stuff works now. This is a summary: Suppose that x,y,z,w are variables that represent complex numbers. In this post I will call any piece of additional information about the values of those variables a constraint. The equalities

$$z=x+iy$$

$$w=x-iy$$

are constraints. This pair of equalities is equivalent to

$$x=\frac{z+w}{2}$$

$$y=\frac{z-w}{2i}$$

These constraints implictly define four maps from ℂ² into ℂ:

$$(x,y)\mapsto z$$

$$(x,y)\mapsto w$$

$$(z,w)\mapsto x$$

$$(z,w)\mapsto y$$

Now we would like to impose one more constraint, $x\in\mathbb R$. This is of course equivalent to $w=z*$. When we do, the maps that are implicitly defined by our constraints change:

$$(x,y)\mapsto z\qquad :\mathbb R^2\rightarrow\mathbb C$$

$$(x,y)\mapsto z^*\qquad :\mathbb R^2\rightarrow\mathbb C$$

$$(z,z^*)\mapsto x\qquad :\{(z,w)\in\mathbb C^2|w=z^*\}\rightarrow\mathbb R$$

$$(z,z^*)\mapsto y\qquad :\{(z,w)\in\mathbb C^2|w=z^*\}\rightarrow\mathbb R$$

Let's call them u,v,F,G respectively. The partial derivatives of u and v are clearly well-defined, and I don't think it's too horrible to write them as

$$\frac{\partial z}{\partial x}=1,\ \frac{\partial z}{\partial y}=i,\ \frac{\partial z^*} {\partial x}=1,\ \frac{\partial z^*}{\partial y}=-i$$

The definition of partial derivative fails miserably for any function $H:\{(z,w)\in\mathbb C^2|w=z^*\}\rightarrow S$, where S is a subset of ℂ, and this of course includes F and G. The "solution" to this "problem" is apparently to define

$$\frac{\partial H(z,z^*)}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)\Big((x,y)\mapsto H(u(x,y),v(x,y))\Big)$$

and similarly for the other partial derivative. This definition is motivated by the fact that if the domain of H had been ℂ², so that the usual definition of partial derivative had worked, $D_1H(u(x,y),v(x,y))$ would have been equal to the right-hand side of the equality above for all $x,y\in\mathbb R$.

So the weird definition of the partial derivatives of a function that's only defined on pairs of the form (z,z*) as a result of the constraint $x,y\in\mathbb R$, is equivalent to just waiting until after we have taken the partial derivatives before we impose that constraint.

What I still don't get about all of this is why we would prefer to make a bunch of weird redefinitions of standard notation and terminology in order to make each step of a nonsensical calculation correct, instead of just saying "hey, let's compute the partial derivative first, and then set w=z*".

Edit: What's even harder to understand is why we would want to describe this result as "z and z* are independent".

 

[Sankaku]

If you don't understand this formula it is only you who is confused.

[ citation needed ]

I referenced 3 books, including 2 you pointed me to. Are you arguing with Ahlfors? Cough up some paper reference where they do the derivation as a limit.


.

 

[Avodyne]

I think A. Neumaier's discussion was clear, and it has helped me to understand this issue much better than I did.

For me, the key point is that partial derivatives with respect a particular variable are well-defined, even if other variables are functions of them.

For example, suppose I have a function g(x,y), where x and y are cartesian coordinates on a plane. The meaning of the partial derivative $\partial g(x,y)/\partial x$ is clear.

Now suppose I am interested in the value of g(x,y) along a curve y=f(x). This is given by g(x,f(x)). But, the partial derivative with respect to x is still well-defined, even though y is no longer "independent" of x.

To be clear, we should write the partial derivative with respect to x in this situation as

[tex]{\partial g(x,y)\over\partial x}\bigg|_{y=f(x)}.[/itex]

Complex derivatives are of this nature, it seems to me. We declare z and z* to be "independent" for purposes of taking partial derivatives, even though we are later going to take z* to be a particular function of z (namely, the complex conjugate).

 

[A. Neumaier]

[ citation needed ]
I referenced 3 books, including 2 you pointed me to. Are you arguing with Ahlfors? Cough up some paper reference where they do the derivation as a limit.

I am not arguing with Ahlfors. I stated a formula which is valid in the Wirtinger calculus, no matter whether or not it is in the book by Ahlfors. One doesn't need a book to see that the limit formula is correct. It follows easily from the other definition. And it can serve as an alternative definition since one can derive from it the formula Ahlfors may have used as definition. (I don't have his book.)

 

[Fredrik]

For example, suppose I have a function g(x,y), where x and y are cartesian coordinates on a plane. The meaning of the partial derivative $\partial g(x,y)/\partial x$ is clear.

Now suppose I am interested in the value of g(x,y) along a curve y=f(x). This is given by g(x,f(x)). But, the partial derivative with respect to x is still well-defined, even though y is no longer "independent" of x.

To be clear, we should write the partial derivative with respect to x in this situation as

[tex]{\partial g(x,y)\over\partial x}\bigg|_{y=f(x)}.[/itex]

I think this notation and terminology is very misleading. The worst part is the notation at the end, but let's start at the beginning. g(x,y) isn't a function. That expression represents a member of the range of the function g. If we write g:ℝ²→ℝ, there's no need to mention coordinates.

If we are only interested in the values of g at points in its domain of the form (x,f(x)), we can consider the restriction of g to the set of such points, but the partial derivatives of that function are undefined. What we need to do here is to define the curve C by C(x)=(x,f(x)) for all x, and to consider the ordinary derivative of g_°C:ℝ→ℝ. I wouldn't describe the fact that what we're really interested in is an ordinary derivative of a different function than the one we started with, as "the partial derivative with respect to x is still well-defined".

Now let's talk about the notation at the end. If $h:\mathbb R^2\rightarrow\mathbb R$ is differentiable, then the partial derivative with respect to the first variable is the function $D_1h:\mathbb R^2\rightarrow\mathbb R$ defined by

$$D_1h(x,y)=\lim_{h\rightarrow 0}\frac{h(x+h,y)-h(x,y)}{h}$$

for all $(x,y)\in\mathbb R^2$. $\partial h/\partial x$ is just an alternative notation for $D_1h$, motivated by the fact that we often use the symbol x as the first variable. The expression

$$\frac{\partial h(x,y)}{\partial x}$$

just means "the value of the function $D_1h$ at (x,y)". So

$$\frac{\partial f(x,g(x))}{\partial x}$$

can only mean $D_1f(x,g(x))$, which is equal to

$$\lim_{h\rightarrow 0}\frac{g(x+h,f(x))-g(x,f(x))}{h},$$

not

$$\lim_{h\rightarrow 0}\frac{g(x+h,f(x+h))-g(x,f(x))}{h}=(f\circ C)'(x)$$

Since $\partial/\partial x$ by definition denotes partial differentiation with respect to the first variable (i.e. exactly the same thing as $D_1$), the expression you used,

$${\partial g(x,y)\over\partial x}\bigg|_{y=f(x)}$$

should therefore be the same thing as

$$D_1g(x,y)\big|_{y=f(x)}$$

and I can only interpret that as "what you get when you replace y with f(x) in the expression $D_1g(x,y)$", and this is $D_1g(x,f(x))$, which is equal to the first of the two limits above, not the second.

 

[A. Neumaier]

You're obviously going to counter by saying that you're talking about a generalization while I was talking about a restriction. This means that you aren't contradicting yourself, but I still find it funny that you're so willing to embrace a redefinition of the term "independent"

My interest in these discussions here on PF is to explain the actual usage of concepts in theoretical physics. One cannot change these traditions, but one can understand them and become confident in their correct use.

In this thread, I was simply explaining in which sense the existing, well-established traditions about df(z^*,z)/dz and ''treating z and z^* as independent'' are fully rigorous and make perfect sense, at least to me.

That you don't like this tradition is a different matter about which I can't argue.

 

[jostpuur]

Suppose we want to solve the equations of motion defined by this Lagrangian.

$$L(\dot{x},\dot{y},x,y) = \frac{1}{2}(\dot{x}^2 + \dot{y}^2) - \frac{C}{2}(x^2 + y^2)$$

The way 1:

$$0 \;=\; D_t \frac{\partial L}{\partial\dot{x}} - \frac{\partial L}{\partial x} \;=\; \ddot{x} + Cx$$
$$0 \;=\; D_t \frac{\partial L}{\partial\dot{y}} - \frac{\partial L}{\partial y} \;=\; \ddot{y} + Cy$$

The way 2:

First we denote $z = x+iy$ and $z^* = x - iy$, and redefine the Lagrangian

$$L = \frac{1}{2}\dot{z}^* \dot{z} - \frac{C}{2} z^* z$$

(Of course not writing $L(\dot{z},\dot{z}^*,z,z^*)$ explicitly.) Then we assume that $z$ and $z^*$ are independent and compute

$$0 = D_t \frac{\partial L}{\partial \dot{z}^*} - \frac{\partial L}{\partial z^*} = \frac{1}{2}\big(\ddot{z} + Cz\big)$$

My question is that why would you use the "way 2"? What do you achieve with it? Is it surely worth all the confusion it will inevitably generate? You could have also obtained the same result by the way 1.

 

[A. Neumaier]

deleted

 

[A. Neumaier]

Suppose we want to solve the equations of motion defined by this Lagrangian.

$$L(\dot{x},\dot{y},x,y) = \frac{1}{2}(\dot{x}^2 + \dot{y}^2) - \frac{C}{2}(x^2 + y^2)$$

My question is that why would you use the "way 2"? What do you achieve with it? Is it surely worth all the confusion it will inevitably generate? You could have also obtained the same result by the way 1.

If the Lagrangian is given in your form, there is no reason to perform the transformation.

But suppose you have a problem where your Hamiltonian is given in the form of an anharmonic oscillator
$$H(z^*,z)= \omega z^*z + g (z^*z)^2 + g' (z^4+(z^*)^4),$$
say. Then you want to write your dynamics directly in terms of z,
$$dz/dt = i dH/dz^*(z^*,z)=\omega z + 2g z(z^*z)^2 + 4g' (z^*)^3,$$
rather than first have to convert it to real and imaginary part, and using the real Hamiltonian equations.

Note that in electrical circuits, say, the variables are naturally given as complex quantities, and the above form is far more natural than the one in terms of real quantities.

 

[jostpuur]

If H is an analytic function of z^* and z then
$$dH(z^*,z)/dz^*=lim_{h\to 0} (H(z^*+h,z)-H(z^*,z))/h$$
makes perfect sense and gives the right result.

I am sorry, I can only say again that z cannot be fixed while you vary $$\bar{z}$$. If you can't see the circular logic in your statement, there is nothing I can do.

In case you haven't seen it: I am fixing both z and z^* and vary a _new_ variable h.
There is nothing circular in my argument; it has a standard, perfectly well-defined, rigorous interpretation.

Neumaier, in the beginning you said that H would be an analytic function of z^* and z, which sounds suspicious, because if H is an analytic function of z, then it is not an analytic function of z^*. It could be that this distracted Sankaku. But I see that what you mean makes sense.

For me, the key point is that partial derivatives with respect a particular variable are well-defined, even if other variables are functions of them.

I see this now too.

Since this has been a confusing thread, it won't hurt if I iterate this a little bit for others to see more explicitly:

If

$$f:\mathbb{C}^2\to\mathbb{C},\quad (z_1,z_2)\mapsto f(z_1,z_2)$$

is a function such that it is complex analytic with respect to the both variables separately, then the following partial derivative functions exists

$$(z_1,z_2) \mapsto (\partial_1 f)(z_1,z_2),\quad (z_1,z_2)\mapsto (\partial_2 f)(z_1,z_2)$$

and it makes sense to use the following notation:

$$\frac{\partial f(z,z^*)}{\partial z} := (\partial_1 f)(z,z^*),\quad \frac{\partial f(z,z^*)}{\partial z^*} := (\partial_2 f)(z,z^*)$$

Sankaku, Frederik, all clear?

My final comment on this is that it's amazing how physicists succeeded in preventing me from understanding this earlier. ;(

 

[jostpuur]

Note that in electrical circuits, say, the variables are naturally given as complex quantities, and the above form is far more natural than the one in terms of real quantities.

Everything is real in classical EM unless something is specifically somehow interpreted as complex. Aren't the complex numbers in electrical circuits only used as a computational trick, because people don't want to deal with formulas

$$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$$
$$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$

but prefer

$$e^{i(A + B)} = e^{iA} e^{iB}$$

instead?

 

[A. Neumaier]

Neumaier, in the beginning you said that H would be an analytic function of z^* and z, which sounds suspicious, because if H is an analytic function of z, then it is not an analytic function of z^*. It could be that this distracted Sankaku. But I see that what you mean makes sense.

Of course. An analytic function in z is as different from an analytic function in z and z^*
as a real function of x and y is different from a real function of x.

If
$$f:\mathbb{C}^2\to\mathbb{C},\quad (z_1,z_2)\mapsto f(z_1,z_2)$$
is a function such that it is complex analytic with respect to the both variables separately, then the following partial derivative functions exists
$$(z_1,z_2) \mapsto (\partial_1 f)(z_1,z_2),\quad (z_1,z_2)\mapsto (\partial_2 f)(z_1,z_2)$$
and it makes sense to use the following notation:
$$\frac{\partial f(z,z^*)}{\partial z} := (\partial_1 f)(z,z^*),\quad \frac{\partial f(z,z^*)}{\partial z^*} := (\partial_2 f)(z,z^*)$$

Yes. And in this case one says that f(z,z^*) is an analytic function of z and z^*.

Note that given an analytic function of z and z^* in the form of a nonanalytic function of z (e.g., f=Re z), one can find out what f(z_1,z_2) must be: The series expansion in powers of z and z^* is well-defined and unique. Replacing in this expansion z by z_1 and z^* by z_2 gives the expansion of f(z_1,z_2).

 

[jostpuur]

"z and z* are independent variables" should mean that we use the notation (z,z*) for points in the domain of the function we're dealing with. This is of course just as trivial, but if * denotes complex conjugation, the domain of the function would have to be (a subset of) the subset of ℂ² that consists of pairs of the form (z,z*), and now we have a problem.

The definition of partial derivative fails miserably for any function $H:\{(z,w)\in\mathbb C^2|w=z^*\}\rightarrow S$, where S is a subset of ℂ, and this of course includes F and G.

These domains emerged from your attempts to guess the meaning for vague statements, but IMO you should forget them now, because they turned out not to be relevant for sensible interpretations of these initially vague statements.

 

[A. Neumaier]

Aren't the complex numbers in electrical circuits only used as a computational trick, because people don't want to deal with formulas

$$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$$
$$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$

but prefer

$$e^{i(A + B)} = e^{iA} e^{iB}$$

instead?

I agree with your statement if you drop the ''only'', which isn't justified for a trick that improves things to an extent that it is virtually everywhere used where physicists work with periodic terms. The latter is far more natural than the former, much easier to remember, much easier to use, and in every respect better behaved.

It is the natural expression of periodicity, as can be seen everywhere: from the Fourier transform or from the solution of linear differential equations with constant coefficients in terms of an eigenvalue problem, from the way the Schroedinger equation is treated. And of course also from the way, linear electrical circuits are analyzed. Once your circuit has more than very few elements, it becomes extremely messy to work with trigonometric functions.

All mathematics consists of tricks to make reasoning shorter. We invent the concept of a prime because it is a useful trick not to have to say each time ''number that is not divisible by any other number apart from one itself'', the decimal notation to be able to say 123 in place of ''one times 100 plus two times ten plus three'', etc. We invent the concept of a phase space vector to be able to abbreviate by the letter x a long list of coordinates. etc. etc..

Mathematics progresses by finding concepts that reduce the labor of precise reasoning to an extent that even very complex matters look comprehensible.

 

[Fredrik]

My question is that why would you use the "way 2"? What do you achieve with it? Is it surely worth all the confusion it will inevitably generate? You could have also obtained the same result by the way 1.

My problem is not so much with "way 2", but with the way it's presented in physics books. If they had said something like

In this Lagrangian, the symbol * doesn't denote complex conjugation, and z* is just another variable. We determine the equations of motion for these two functions, and find a) that they're exactly the same, and b) that the complex conjugate of any solution is a solution. This means that if we set z* equal to the complex conjugate of z after we have determined the equation satisfied by both, we obtain a theory of a single complex-valued function instead of a theory of two.​

I would have been OK with it.

Neumaier, in the beginning you said that H would be an analytic function of z^* and z, which sounds suspicious, because if H is an analytic function of z, then it is not an analytic function of z^*. It could be that this distracted Sankaku. But I see that what you mean makes sense.

It definitely distracted me. I didn't even look at the limit right away. But I agree that the limit is well-defined for all values of z, assuming that H is defined and analytic in an open set that contains the set of pairs of the form (z*,z). This is just the standard (not Wirtinger) definition of partial derivative.

With that in mind, the definition

$$\frac{\partial H(z^*,z)}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)\Big((x,y)\mapsto H(x-iy,x+iy)\Big)$$

seems strange and unnecessary. This is the sort of stuff that the Wikipedia article put into my head. Maybe it's useful for something, but I don't think we need it here.

These domains emerged from your attempts to guess the meaning for vague statements, but IMO you should forget them now, because they turned out not to be relevant for sensible interpretations of these initially vague statements.

I agree that we have no interest in any functions with that domain. All we need to do is to wait until after we have found the partial derivative (which is another function from ℂ² into ℂ) until we set z*=(complex conjugate of z).

My final comment on this is that it's amazing how physicists succeeded in preventing me from understanding this earlier. ;(

I have felt that way many times. I could totally understand the frustration you displayed in #6. I still get angry when I think about how tensors were explained to me in 1994.

 

[Avodyne]

My problem is not so much with "way 2", but with the way it's presented in physics books. If they had said something like

In this Lagrangian, the symbol * doesn't denote complex conjugation, and z* is just another variable. We determine the equations of motion for these two functions, and find a) that they're exactly the same, and b) that the complex conjugate of any solution is a solution. This means that if we set z* equal to the complex conjugate of z after we have determined the equation satisfied by both, we obtain a theory of a single complex-valued function instead of a theory of two.​

I would have been OK with it.

They don't say this because that's not what's done.

Suppose z=x+iy is a complex field, and z*=x-iy is its complex conjugate. We can express the lagrangian as a function of x and y, or as a function of z and z*. I will call the first function R and the second function C. (R is to remind us of "real" and C of "complex".) R is a function from ℝ² into ℝ, and C is a function from ℂ² into ℂ. These functions are related by

R(x,y)=C(x+iy,x-iy).

From this, it follows immediately (using the chain rule) that the derivatives of these functions are related by (using your non-standard notation)

D₁R = (D₁ + D₂)C

D₂R = (1/i)(D₁ - D₂)C.

We can now solve for D₁C and D₂C, with the result

D₁C = (1/2)(D₁ + iD₂)R

D₂C = (1/2)(D₁ - iD₂)R.

(These are the formulas that you call "strange and unnecessary".)

The equations of motion are

D₁R = 0

D₂R = 0.

Using the "strange and unnecessary" formulas, we find that

D₁C = 0

D₂C = 0.

Now it so happens that it is generally easier to compute D₁C and D₂C than it is to compute D₁R and D₂R. One reason for this is that C has the property that C(x+iy,x-iy) must be real, and this implies that D₂C=(D₁C)^* when evaluated at (x+iy,x-iy). Thus we only need to compute D₁C.

Note that it is never necessary to say that "the symbol * doesn't denote complex conjugation".

 

[Fredrik]

Note that it is never necessary to say that "the symbol * doesn't denote complex conjugation".

OK, you're right. I have no objections to anything you did in this post.

your non-standard notation

I think all of these notations are standard, but perhaps I picked the least popular one:

$$D_1f(x,y)=\partial_1 f(x,y)=f_{,1}(x,y)=\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x_1}\bigg|_{(x,y)}f(x_1,x_2)$$

Edit: I think you actually made the best post in the entire thread. It explains a lot, without any weird terminology or strange definitions. One of the things you proved is that there's no need to take the formulas I called "strange and unnecessary" as definitions, because the result

$$\frac{\partial C(z,z^*)}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)C(x+iy,x-iy)$$

follows from the standard definition of a partial derivative, assuming of course that the function C is defined and analytic in an open subset of ℂ² that includes the point (z,z*).

 

[Avodyne]

OK, you're right. I have no objections to anything you did in this post.

!

I think you actually made the best post in the entire thread.

! !

I think all of these notations are standard, but perhaps I picked the least popular one:

Well, I haven't seen it before (except in Mathematica). But if the arguments were called $(x_1,x_2)$, then I would know what $\partial_1$ meant.

Incidentally, the property that C(x+iy,x-iy) must be real is better expressed as the statement that C must be a symmetric function on ℂ².

 

[A. Neumaier]

OK, you're right. I have no objections to anything you did in this post.I think all of these notations are standard, but perhaps I picked the least popular one:

$$D_1f(x,y)=\partial_1 f(x,y)=f_{,1}(x,y)=\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x_1}\bigg|_{(x,y)}f(x_1,x_2)$$

Edit: I think you actually made the best post in the entire thread. It explains a lot, without any weird terminology or strange definitions. One of the things you proved is that there's no need to take the formulas I called "strange and unnecessary" as definitions, because the result

$$\frac{\partial C(z,z^*)}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)C(x+iy,x-iy)$$

follows from the standard definition of a partial derivative, assuming of course that the function C is defined and analytic in an open subset of ℂ² that includes the point (z,z*).

The point of the Wirtinger calculus is that this assumption (which I made for didactical reasons) is not needed when you work in terms of his definitions. Because in general you only have a function f(z) that is continuous in z and continuously differentiable in Re z and I am z (i.e., considered as a function in R^2), such as F(z)=|Re z|^3-|Im z|^3. One cannot apply Avodyne's recipe in that case, but the Wirtinger derivative 9and the second derivatives) are still well-defined.

It is a bit like the difference between a real analytic function and a real differentiable function. To get the latter, you need more careful definitions than to get the former.

 

[A. Neumaier]

Incidentally, the property that C(x+iy,x-iy) must be real is better expressed as the statement that C must be a symmetric function on ℂ².

But this is not a requirement for the calculus to work.

 

[Fredrik]

!

! !

Do I seen so stubborn and grumpy that I can't give someone else credit for something good? Maybe it's the avatar.

Well, I haven't seen it before (except in Mathematica).

You made me wonder if the book that I picked up the D_if notation from (a little known Swedish book) was using a non-standard notation, but I just checked my copy of the infamous "baby Rudin" (Principles of mathematical analysis), and it's the default notation there too.

I actually like the notation $f_{,\,i}$ a lot, because it makes the chain rule look so nice, e.g.

$$(f\circ g)_{,i}(x)=f_{,j}(g(x))g_{j,i}(x)$$

 

[A. Neumaier]

I actually like the notation $f_{,\,i}$ a lot, because it makes the chain rule look so nice, e.g.
$$(f\circ g)_{,i}(x)=f_{,j}(g(x))g_{j,i}(x)$$

The notation without indices (regarding D as a vector with components D_i) looks even nicer: D(f(g(x)) = Df(g(x))Dg(x)

 

[Avodyne]

Arrggh! I somehow managed to screw up some minus signs in my big post above, and now I'm not seeing the "edit" button.

Correct versions are

D₁R = (D₁ + D₂)C

D₂R = i(D₁ - D₂)C.

We can now solve for D₁C and D₂C, with the result

D₁C = (1/2)(D₁ - iD₂)R

D₂C = (1/2)(D₁ + iD₂)R.

This implies

$${\partial\over\partial z}C(z,z^*)={1\over2}\left({\partial\over\partial x}-i{\partial\over\partial y}\right)C(x+iy,x-iy)$$

which is the same formula given by A.Neumaier in post #17.

 

[Avodyne]

in general you only have a function f(z) that is continuous in z and continuously differentiable in Re z and I am z (i.e., considered as a function in R^2), such as F(z)=|Re z|^3-|Im z|^3. One cannot apply Avodyne's recipe in that case

I don't see the problem ... Is it because of the absolute-value signs?

 

[Avodyne]

Do I seen so stubborn and grumpy that I can't give someone else credit for something good? Maybe it's the avatar.

I thought my math would be too "lowbrow" for you, though I did make an effort to specify the domain and range of my functions, and to avoid sloppy language, such as referring to the function R as R(x,y).

 

[A. Neumaier]

I don't see the problem ... Is it because of the absolute-value signs?

Yes. The absolute value is not an analytic function. i.e., you can't expand it into a power series.