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What is the meaning of having $\bar{z}$ in your description of a function, i.e. $f(z,\bar{z})$? The conjugate is simply a function of the complex number, z, so I don't understand why you need to reference it as a second variable. It seems equivalent to noting $x^2$ as a variable in some equation like $f(x,x^2)=x+x^2$. Why does this happen?

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    $\begingroup$ Perhaps it's just to really emphasize the relative dependence on $z$ and its conjugate. Since you can frequently think of an analytic function as a "pure function of $z$," this is a much more meaningful consideration than $f(x, x^2)$ would be. $\endgroup$ – user61527 Jul 1 '14 at 7:34
  • $\begingroup$ If $f$ is a function of two complex variables, like say $f(z,w) = z w^3$, it makes perfect sense to talk of $f(z,\bar{z})$. For example if $z=i$ you would get $f(z,\bar{z})=f(i,-i)=i(-i)^3=-1$. $\endgroup$ – Georges Elencwajg Jul 1 '14 at 7:52
  • $\begingroup$ @GeorgesElencwajg Certainly, but I mean specifically a function of z and it's complement. $\endgroup$ – user82004 Jul 1 '14 at 7:56
  • $\begingroup$ Dear Anthony: you seem not to understand my comment. Could you give an example of a function of the form $f(z,\bar z)$ ? By the way $\bar z$ is called the conjugate of $z$, not its complement. $\endgroup$ – Georges Elencwajg Jul 1 '14 at 8:07
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    $\begingroup$ The point is to make it explicitly clear that $f$ is not holomorphic. But I agree with Georges; it's poor notation. $\endgroup$ – user98602 Jul 1 '14 at 9:42
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"variable" is a dangerous word to throw around.

In mathematics, strictly speaking we don't need to name the arguments of the functions, and a function has only one argument. Every function is a function from a set $A$ to a set $B$ and this description doesnt talk about variables.

Now in some cases (all the time in physics) we want to talk about functions of several variables, and more importantly, we want to do change of variables, without changing the name of the function.

For example, if you drop a rock into a well and you consider its speed $s$, altitude $z$, and acceleration $a$, any of those quantities is technically a function of any of the other, so we want to write things like $s(z), a(z), a(s), \ldots$ but where the two $a$ denotes, mathematically, different functions.

When dealing with functions with multiple variables, things get worse when you consider partial derivatives. For example if temperature $T$ of a gas is a function of volume $V$ and pression $P$, but the size and the pression of the container changes with time $t$, you may consider the $T$ as a function of $t$ and $P$. Now the funny thing is that $dT/dP$ isn't the same thing wether you are talking about the function $T(P,V)$ or $T(P,t)$.

Back to your question, functions from $f : \Bbb C$ to $\Bbb C$ are often talked about as functions $\Bbb R^2$ to $\Bbb C$ with two real arguments $x,y$.

When you do a real change of variable like $x' = (x+y)/\sqrt 2, y' = (x-y)/\sqrt 2$, the two new variables have nothing to do with each other and in this example the change of variable is a mere rotation of the plane.

Say we want to make the $\Bbb C$-linear change of variables $z = x+iy, \bar z = x-iy$ (or in the other direction $x = (z+\bar z)/2, y = i(\bar z-z)/2$) ($z$ and $\bar z$ are just name)s.

Here $z$ takes complex values and $\bar z$ is the conjugate of $z$ when $x$ and $y$ are real. Note that for the change of variable to make sense, we are plugging something syntaxically complex ($i(z-\bar z)/2$) into $y$.
So maybe everything makes better sense if instead of saying that $x$ and $y$ are real variables, we consider them as complex variables (we will note this function $\hat f_1 : \Bbb C^2 \to \Bbb C$, with arguments $x$ and $y$). Because then, the change of variables is almost like a "complex rotation" (we call the resulting function $\hat f_2 : \Bbb C^2 \to \Bbb C$, with arguments $z$ and $\bar z$).
And this time, $z$ and $\bar z$ are two complex variables of $\hat f_2$ that have nothing to do with each other, just like $x$ and $y$ for $\hat f_1$. It just happens that $\bar z$ is the conjugate of $z$ exactly when $x$ and $y$ are real.

You can then talk about $d\hat f_2/dz$ and $d\hat f_2/d\bar z$ and the following are equivalent :

1) the original function $f: \Bbb C \to \Bbb C$ is holomorphic
2) $d\hat f_2/d\bar z = 0$
3) $\hat f_2(z,\bar z)$ only depends on $z$ and not on $\bar z$
4) $i d\hat f_1/dx = d\hat f_1/dy$
5) $\hat f_1(x,y)$ only depends on the value of $x+iy$ and not on the particular (possibly complex) values of $x$ and $y$.

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$f$ is a function of two variables:

$$f:\mathbb C^2\to\mathbb C$$ $$f(x, y) =\ ...$$

We then compose it with the functions $z\to z$ and $z\to\overline z$ to obtain a function of one variable.

$$f(z, \overline z): \mathbb C\to\mathbb C$$

I suppose why this is useful depends on the context.

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I don't think this was a good answer, but other peoples' comments are better than the answer so I'm not deleting it.

The function $x + x^2$ is easily expressed as a function of $x$, but how would you express $z + \bar z^2$ as a function of z ?

You could do it using functions $Re$ and $Im$ which project the real and imaginary parts and get $z + Re(z)^2 + Im(z)^2 -2iRe(z)Im(z)$, but it's clumsy isn't it.

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  • $\begingroup$ I understand that it's clumsy, but is there really no similar disparity with real functions? For instance, is it allowed to put an $|x|$ in a formula? It just seems like you lose more by putting $\bar{z}$ as a separate variable... $\endgroup$ – user82004 Jul 1 '14 at 7:38
  • $\begingroup$ You can put $|x|$ as a variable in a function, e.g. $y = |x| $ gives a V shaped graph. You can also have real functions of more than one variable where the variables are related through a functional dependency of their own, e.g. in the Hamiltonian expression of mechanics. $\endgroup$ – Tom Collinge Jul 1 '14 at 7:43
  • $\begingroup$ Dear Tom, contrary to what you claim $z + \bar z^2$ is is already expressed as a perfectly legitimate function of $z$, albeit not holomorphic but nevertheless $\mathcal C^\infty$. If necessary look at the definition of "function" in set theory. $\endgroup$ – Georges Elencwajg Jul 1 '14 at 7:49
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    $\begingroup$ Dear Anthony, the expression $z + \bar z^2$ is a perfectly clear definition of a function of $z$ and there is absolutely no need to write it in a roundabout way as a function of $Im(z), Re(z)$. On the contrary: writing $z + \bar z^2$ immediately shows that the function is not holomorphic whereas the clumsy way does not, and forces you to use the Cauchy-Riemann equation. $\endgroup$ – Georges Elencwajg Jul 1 '14 at 8:03
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    $\begingroup$ Dear Tom, your gracious sentence at the beginning of your edited answer displays great modesty, self-criticism and honesty. I value this much more than any technical expertise on a question of notation. Bravo and +1. $\endgroup$ – Georges Elencwajg Jul 1 '14 at 10:17
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Recall that one form of the Cauchy Riemann equations is given by $\frac{\partial }{\partial\overline{z}} = 0$.

By writing $f(z,\overline{z})$ we imply that that $\frac{\partial f}{\partial\overline{z}} \ne 0$ and make it clear that $f$ is definitely not analytic.

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Consider a function of $z$, say $g(z) = \bar z + z^2$. It is a function fo $x=\Re z, y = \Im z$: $$g(x,y) = x-iy + x^2-y^2 + 2ixy$$

Now, as $x = \frac12 (z+\bar z), y = \frac 1{2i}(z-\bar z)$ it is without confusion a function of $(z,\bar z)$ (without "impure" operators like $\bar {z}$)

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What are independent variables?

Let me begin with an example from differential equations: $y''+y'=0$. To solve this, we set $v=y'$ and reduce to the first order problem $v'+v=0$ hence $v=c_1e^{-t}$. Continuing, $v=y'$ hence $c_1e^{-t}=y'$ so we integrate to obtain $y=c_2+c_1e^{-t}$. Notice, the differential equation $y''+y'=0$ is often written symbolically as a general object by $F(y,y',y'';t)=0$. We list the dependent variable $y$ as well as the derivatives of $y$ up to the highest order which appears in a nontrivial fashion in the given ODE.

Should we instead write $F(y,t)=0$? I think not. Of course, $y',y'', \dots$ can be obtained via differentiation, but, the listing of those variables as if they are independent has merit. Furthermore, we can make them genuinely independent by lifting to the jet space.

When we study variational calculus in physics, similar comments apply. We make distinctions between writing something as a function of $p$ verses $\dot{q}$ even though, $p = m\dot{q}$. It's the difference between a Hamiltonian and a Lagrangian formulation.

My point is simply this, when independence is indicated between objects which appear to have dependence it is often a short-hand for an abstraction. I wouldn't dismiss it as a mistake. Rather, it is an invitation to take a different view of the problem.

That said, concerning $z$ and $\bar{z}$ I have yet to clearly understand the analog of jet space. As far as I can cipher, the $z$ and $\bar{z}$ notation is simply a complex notation for functions of real variables. It does allow beautiful generalizations of well-known complex formulae. For example, see Pompeiu's formula as a generalization of Cauchy's integral formula. Or for more background, we ought to read about Wirtinger derivatives.

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