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I am quite confused about how to understand $\frac{\partial f}{\partial z}f(z,\bar z).$ Do $z$ and $\bar z$ in $f(z,\bar z)$ act the same way as $x$ and $y$ in $f(x,y)$?

If so, how can we prove this?

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More context would help. I think it would only make sense where $f$ is holomorphic in its first argument and anti-holomorphic in its second. (But no, $z$ and $\bar{z}$ are not independent.) –  anon Apr 5 '12 at 4:39
Actually, that might be a sufficient but not necessary condition for well-definedness. Perhaps more exotic $\mathbb{C}^2\to\mathbb{C}$ functions exist where $\large \frac{\partial}{\partial z}\normalsize f(z,\bar{z})$ makes sense? –  anon Apr 5 '12 at 4:44
You might find the answers to this question helpful. –  robjohn Apr 5 '12 at 8:54
@robjohn. The first sentence in your link is spot on: "The nomenclature of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ is confusing because it gives the impression that these are really partial derivatives with respect to two independent variables, $z$ and $\bar{z}$. However, it is clear that $z$ and $\bar{z}$ are not independent. " –  Georges Elencwajg Apr 5 '12 at 9:47
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Given an open subset $U\subset \mathbb C=\mathbb R^2$ and a $\mathcal C^\infty$-function $f:U\to \mathbb C$, one defines
$$ \frac{\partial f}{\partial z} = \frac{1}{2}\left(\frac{\partial f}{\partial x} -i\frac{\partial f}{\partial y}\right) \in \mathcal C^\infty (U) \; \text {and} \frac{\partial f}{\partial \bar z} = \frac{1}{2}\left(\frac{\partial f}{\partial x} +i\frac{\partial f}{\partial y}\right)\in \mathcal C^\infty (U)$$Notice carefully that I wrote $\mathbb C=\mathbb R^2$, an equality not an isomorphism: one has the same set but the notation $\mathbb C$ means we have endowed $\mathbb R^2$ with with its well-known field structure. Consequently one also writes $z=x+iy=(x,y)$.

This is essentially all there is to say. No mystery here: we have just defined two differential operators $ \frac{\partial f}{\partial z}, \frac{\partial f}{\partial \bar z} \in Der _{\mathbb C} (\mathcal C^\infty (U)) $.
There is an analogous punctual version $ \frac{\partial }{\partial z}\mid _{z_0}, \frac{\partial }{\partial \bar z}\mid _{z_0} \in Der _{\mathbb C} (\mathcal C^\infty _{z_0},\mathbb C) $ for functions defined only on a neighbourhood of a fixed point $z_o\in \mathbb C$
[$Der$ stands for derivation, a fancy algebraization of the good old Leibniz rule for taking the derivative of a product]

For example we have $\frac{\partial (x^2)}{\partial z}=x,\quad \frac{\partial (\sin xy +i e^ x)}{\partial \bar z}=\frac{1}{2}[y\cos xy+i(e^x+x\cos xy)]$

As I am sure you know, a $\mathcal C^{\infty} $ function $f$ is holomorphic iff $\frac{\partial f}{\partial \bar z}=0 $ and in that case $f'(z)=\frac{\partial f}{\partial x}(z)$.
For example if $f(z)=z^2=x^2-y^2+2ixy$ then $f'(z)=2z=2x+2iy=\frac{\partial (x^2-y^2+2ixy)}{\partial x}$

And what about $\frac{\partial }{\partial z}f(z,\bar z)$ ? Forget about that notation : it makes absolutely no sense if $z$ is not real, because already $f(z,\bar z)$ is absolutely not defined !
[Actually, there are contorsions which define $f(z,\bar z)$ for real-analytic functions like polynomials, but they are artificial, hide the utmost simplicity of the Wirtinger calculus (that's the name of the guy who introduced the partials $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \bar z}$) and thus should be avoided]
In the same vein, $z$ and $\bar z$ are not at all independent. Quite the contrary: $\bar z$ is completely determined by $z $ !
What people mean when they use that ridiculous phrase is probably that $\frac{\partial z}{\partial \bar z}=0$, but then they should say just that and not introduce this absurd terminology of "independent variables".
[I have only noticed now that this was your actual question! Sorry for that.]

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Consider a function $f:\mathbb C\rightarrow \mathbb C$, use the identification $\mathbb C\simeq\mathbb R^2$ to write $$ f(x+iy)=(u(x,y),v(x,y)), $$ and assume it is differentiable w.r.t to $x$ and $y$. By definition, $$ \frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} -i\frac{\partial}{\partial y}\right),\qquad \frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} +i\frac{\partial}{\partial y}\right), $$ which means that, if $z_0=x_0+iy_0$, $$ \frac{\partial}{\partial z}f(z_0)=\frac{1}{2}\left(\frac{\partial}{\partial x}u(x_0,y_0)+\frac{\partial}{\partial y}v(x_0,y_0),\frac{\partial}{\partial x}v(x_0,y_0)-\frac{\partial}{\partial y}u(x_0,y_0)\right) $$ and $$ \frac{\partial}{\partial \bar z}f(z_0)=\frac{1}{2}\left(\frac{\partial}{\partial x}u(x_0,y_0)-\frac{\partial}{\partial y}v(x_0,y_0),\frac{\partial}{\partial x}v(x_0,y_0)+\frac{\partial}{\partial y}u(x_0,y_0)\right). $$ From this, you may check that $$ \frac{\partial}{\partial z}\bar z=0,\qquad \frac{\partial}{\partial \bar z}z=0, $$ which shows the "independence" between $z$ and $\bar z$ to be similar than the one between $x$ and $y$ you were looking for.

These derivatives are pretty convenient to manipulate, and note that if $f$ is holomorphic, namely satisfies the Cauchy-Riemann equations $$ \frac{\partial}{\partial x}u(x_0,y_0)=\frac{\partial}{\partial y}v(x_0,y_0),\quad \frac{\partial}{\partial x}v(x_0,y_0)=-\frac{\partial}{\partial y}u(x_0,y_0), $$ then $$ \frac{\partial}{\partial z}f(z_0)=\left(\frac{\partial}{\partial x}u(x_0,y_0),\frac{\partial}{\partial y}v(x_0,y_0)\right),\qquad \frac{\partial}{\partial \bar z}f(z_0)=0. $$

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Applying $\partial/\partial z$ only works when $f$ is holomorphic, and the formula for it in terms of $\partial/\partial x$ and $\partial /\partial y$ works when the Cauchy-Riemann equations are satisfied. A quick check shows that $h(z)=\bar{z}$ does not satisfy CR and thus $\partial \bar{z}/\partial z$ does not exist (similarly, by symmetry, $z$ is not anti-holomorphic). | Also: The question is critically about functions of two complex variables. Certainly this discussion is relevant background but does not really count as progress in my opinion. –  anon Apr 5 '12 at 5:55
@anon : No, the question is about functions of one complex variable. $\partial/\partial z$ applies to any $\mathbb R^2$-differentiable complex functions, see the definition above, I don't see what you refer to. –  Student Apr 5 '12 at 6:03
How is $f(x+yi,r+si)$ a function of one complex variable? And what you refer to appear to be the Wirtinger derivatives, which are more generally defined than the usual complex derivatives under their usual definition (I think this is worth note). –  anon Apr 5 '12 at 6:07
Yes, they are indeed the Wirtinger derivatives, and your link presents the same definitions that I've described above. Nobody speaks about $f(x+iy,u+iv)$, but of $f(x,y)=u(x,y)+iv(x,y)$ : $f$ is a function taking values in $\mathbb C$. –  Student Apr 5 '12 at 6:18
Student, you read the question right? It's clear $z$ is a complex variable (otherwise why use a letter for complex variables, and more importantly why write $\bar{z}$ at all?), thus each argument in $f(\cdot,\cdot)$ is complex. (Actually, now that I see OP also wrote $f(x,y)$, I'm not sure I know what he's talking about precisely... Edit: scratch that, he's talking about an analogy with the components of $\mathbb{R}^2$ being independent of each other (i.e. one can be changed while the other is fixed).) –  anon Apr 5 '12 at 6:21
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All this makes perfect sense; you just have to use the right definitions:

Let $f: \mathbb{C}\simeq \mathbb{R}^2\to \mathbb{C}$ be real analytic, that is, if you think of it as being defined on $\mathbb{R}^2$ it has a power series expansion in $(x,y)$ at every point in $\mathbb{R}^2$ (note that this is weaker than being holomorphic). Then $f$ can be extended to a holomorphic function $D\to\mathbb{C}$, where $D\subset \mathbb{C}^2$ is a neighborhood of $\mathbb{R}^2 \subset \mathbb{C^2}$, so you can write $f(x,y)$ where $x,y$ are now complex! Now consider the following change of variables: $$(z,\bar z)=(x+iy,x-iy)$$ Here we regard $\bar z$ as just a symbol that is independent of $z$. Then it makes sense to write $f(z,\bar z)$; $\partial_z$, $\partial_{\bar z}$ are the Wirtinger derivatives, and $\partial_{\bar z}f=0$ (that is, $f$ is actually independent of $\bar z$) is equivalent to $f$ being holomorphic.

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Dear Florian, could you please explain how $\bar z$ is "a symbol that is independent of $z$" and at the same time write that when $z=x+iy$, then $\bar z=x-iy$ ? This says that $\bar z$ is a function of $z$, i.e. completely depends on $z$ by the very definition of "function" . –  Georges Elencwajg Apr 5 '12 at 9:30
OK, $\bar z$ is just a symbol means that it's unrelated to $z$, so if this confuses you you could call it $w$ and say $(z,w)=(x+iy,x-iy)$ is the new coordinate system (it's a linear coordinate system, that is, just another basis of $\mathbb{C}^2$). And it's not true that "when $z=x+iy$ then $\bar z=x-iy$" because the decomposition of $z$ is not unique -- remember that $x$ and $y$ are allowed to be complex. –  Florian Apr 5 '12 at 12:48
Dear Florian, I still don't understand what you are doing. If now $x,y$ are allowed to be complex numbers , what do you you call "the following change of variables" ? For example if $z=1+i$ and $\bar z=3-4i$ what are $x$ and $y$ ? Also, a linear coordinate system on $\mathbb C^2$ is not a basis of $\mathbb C^2$, but a pair of linearly independent linear forms. –  Georges Elencwajg Apr 5 '12 at 13:09
If $z=1+i$, $\bar z=3-4i$ then $x=2-\frac{3}{2}i$, $y=\frac{5}{2}+i$. More generally, the inverse transformation is $x=\frac{1}{2}(z+\bar z)$, $y=\frac{i}{2}(\bar z - z)$. –  Florian Apr 5 '12 at 14:13
Ah, I see: thanks. –  Georges Elencwajg Apr 5 '12 at 14:45
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