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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.)2012-04-05
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    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?2012-04-05
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    You might find the answers to [this question](http://math.stackexchange.com/questions/85648/why-can-the-complex-conjugate-of-a-variable-be-treated-as-a-constant-when-differ) helpful.2012-04-05
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    @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. "2012-04-05

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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]
Edit
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.2012-04-05
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    @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.2012-04-05
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    How is $f(x+yi,r+si)$ a function of one complex variable? And what you refer to appear to be the [Wirtinger derivatives](http://en.wikipedia.org/wiki/Wirtinger_derivatives#Functions_of_one_complex_variable), which are more generally defined than the usual complex derivatives under their usual definition (I think this is worth note).2012-04-05
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    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 $\mathbb C$.2012-04-05
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    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).)2012-04-05
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    Don't you think that $\phi:\mathbb C\rightarrow\mathbb C$, $\phi(z)=\bar z$ is a function of $z$ ? You may then write $g(z)=f(z,\phi(z))$ which is a function of one complex variable. $z$ and $\bar z$ are NOT independent variables, it is just classical and convenient to write $f(z,\bar z)$ once you use the Wirtinger derivatives.2012-04-05
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    Yes, and $f(g(x),h(x))$ is also a function of one variable, but $f(\cdot,\cdot)$ itself a function of two. If you're asking for what $f$ allows $f(g(x),h(x))$ to have so-and-so properties for given $g,h$, you are posing a question about the space of functions of two variables. | "It is classical and conveninent to write $f(z,\bar{z})$ once you use the Wirtinger derivatives." Could you elaborate on this? It seems that *this* would actually answer OPs question, if I understand correctly.2012-04-05
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    Indeed, but you can't write $y=h(x)$. I understand the OP question as : "in $\mathbb R^2$, you can see $x$ and $y$ as two different variables, and $(\partial/\partial x)y=(\partial/\partial y)x=0$. Then, you add the complex structure, look at $f:\mathbb C\rightarrow\mathbb C$, and keep in mind that it is a function of $(x,y)\in\mathbb R^2$ too. After that, two type of derivations are introduced, $\partial/\partial z$ and $\partial/\partial \bar z$, which should not be confused with $d/dz$, and it looks like $z$ and $\bar z$ are independent as $x$ and $y$ are.2012-04-05
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    It is clearly not the case since $\bar z$ is a function of $z$, but there is still an independence at the level of the differential operators. The notation $f(z,\bar z)$ could create ambiguities, I just claim they are pretty convenient.2012-04-05
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    Okay, so we write $f(z,\bar{z})$ for a function $f$ of *one* complex variable *by convention*, even though it's the same as the convention for a function of two arguments almost everywhere else in math? That's my interpretation of your "classical and convenient" comment, though you haven't explicitly said so (and I am completely unfamiliar with this area of complex analysis). If this is the case, why didn't you just say that to begin with instead of letting me have a fit? :P2012-04-05
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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" .2012-04-05
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    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.2012-04-05
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    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.2012-04-05
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    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 $x=\frac{1}{2}(z+\bar z)$, $y=\frac{i}{2}(\bar z - z)$.2012-04-05
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    Ah, I see: thanks.2012-04-05
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    I knew abour Wirtinger derivatives defined just formally as $\frac{1}{2} (\partial _x \pm i \partial_y)$ on complex functions on complex plane for a long time. But this post is first truly satisfying explanation of some manipulations with functions of $z, \bar z$ one can see e.g. in physics textbooks. Thanks!2016-07-06