How does a metric of the form $\mathrm{d}z \mathrm{d}\bar z$ work, if $z$ and $\bar z$ are not independent? My question is motivated by 2D CFT where one works in "complex coordinates". The question is the following:
Suppose I am in 2D flat Euclidean space, i.e.
$$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$$
One of the crucial things here is that $x$ and $y$ are independent. Now it well known that the plane can also be described by one complex number. The point here is that it is ONE complex number, so once I write down:
$$z = x+iy$$
I have everything to describe the complex plane. And yet in every book, they then define
$$\bar z = x-iy$$
so the complex conjugate and write down the metric:
$$\mathrm{d}s^2 = \mathrm{d}z \mathrm{d}\bar z$$
Of course, the norm in the complex plane is given by $z\bar z$ but what I don't understand is that the complex conjugate is not an independent variable, so how does the above metric make sense. Of course, often in the books, they at some point mention that they complexify and essentially consider $\mathbb{C}^2$ in which case $z$ and $\bar z$ are to be considered as independent, then the metric makes more sense. But if you don't complexify, how am I supposed to understand the said metric?
Edit: It seems that the standard answer is that you consider the complexified $\mathbb{C}^2$, but that wasn't my question: If I don't want to go to $\mathbb{C}^2$, and just consider $\mathbb{R}^2$ as $\mathbb{C}$, what would the metric be?
 A: The deal in conformal field theory is the following.
One starts from a two-dimensional conformal field theory written in terms of the variables $x_0, x_1$ and realises that this can be re-written in terms of the chiral light-cone variables $x^{\pm} = x_0\pm x_1$. The metric $ds^2 = d{x_0}^2 + d{x_1}^2$ can equivalently be re-written as $ds^2 = dx^+ dx^-$. The theory is still, so far, two-dimensional.
It is convenient at this point to introduce the below Cayley transformation:
$$
C\colon\mathbb{R}\to S^1\setminus \{-1\},\qquad x^{\pm}\mapsto \frac{1\pm ix^{\pm}}{1\mp ix^{\pm}} = z(\bar{z})
$$
where the new $z$ variable lives on a circle, i. e. on a one-dimensional manifold. The initial two-dimensional theory described on the plane via $x^0, x^1$ has now been mapped onto two one-dimensional circles, hence one two-dimensional copy has been divided into two one-dimensional copies. This is usually referred to as compactification of the complex plane onto two circles (minus one point). The metric $ds^2 = dz d\bar{z}$ is a plane change of coordinate, no particular meaning.
Fields depending on the chiral light cone coordinates are mapped into the tensor product of two algebras, namely 
$$
\phi(x^+, x^-)\mapsto \phi_+(x^-)\otimes 1_- \pm 1_+\otimes \phi_-(x^+).
$$
Conformal transformations are now those transformations $f$ only involving dependencies on $z, \bar{z}$ singularly, without mixing them up. This translated into the fact that each Cayley circle is mapped onto itself without intersecting the other (and therefore the two field algebras remain separated), in particular diffeomorphisms of the circle do the job.
The deal to buy is, so to speak, that $z, \bar{z}$ represent coordinates on a circle, rather than the standard expressions $x\pm iy$ (sometimes they are even re-written as $e^{2(x_0 \pm i x_1)}$).
A: This is actually a more general question--I think--which came up a few times when I was teaching complex analysis: the question was, how do we understand $$\int_{z_0}^{z_1} dz~f(z)$$ when $z$ is a complex variable? And the answer is, we have to draw a curve $C$ on the plane $\mathbb C$ that we want to integrate over, some $z(t) \in (\mathbb R\to\mathbb C)$ for some parameter parameter $t\in(0,1)$ such that $z(0,1) = z_{0,1}.$ The integral is m then unambiguously both$$\int_Cdz~f(z)=\int_0^1dt~f(z(t))~z'(t).$$Usually after a few examples they got the general idea.
Similarly, for you with this metric, probably to get an actual distance requires that you compute like$$\int_C\sqrt{dz~d\bar z} = \int_0^1 dt~|z'(t)|.$$
