TITLE: Independence of $z$ and $\bar{z}$
QUESTION [6 upvotes]: I am teaching complex analysis to physics students next semester and I would like to discuss the fact that $z$ and $\bar{z}$ are functionally(?)/algebraically(?) (what is the correct terminology, anyway?) independent as early as possible in a course. Ideally, after I introduce complex numbers $z$ as a pair of real numbers and develop the basic complex algebra, I would like to define the complex conjugate $\bar{z}$ and then prove that $\bar{z}$ cannot be obtained from $z$ using algebra, i.e. using $+,-,\times,\div$. Now, my question is, what is the strongest statement of independence of $z$ and $\bar{z}$ that I can prove using basic complex algebra?
Rephrasing the question: how can I prove that $z$ and $\bar{z}$ are algebraically independent in the ordinary sense of the word? I want to prove there exists no nontrivial polynomial $P_n(z,\bar{z})$ with complex constants so that $P_n(z,\bar{z}) = 0$ for any $n \in \mathbb{N}$ (including $n \to \infty$).

REPLY [2 votes]: The difficulties arise because we use the same letter $z$ to denote some individual complex number (and then $\bar z$ is the complex conjugate of $z$), and (b) as name for the identity function on ${\mathbb C}$. In the second interpretation $\bar z$ is some other complexvalued function on ${\mathbb C}$. You now ask whether this other function might be analytically related to the function $z$. In other words: Is there an analytic function $f$ (polynomial, algebraic expression, or more complicated), defined at least in some neighborhood $U$ of the origin, such that
$$\bar z =f(z)\qquad(z\in U)\quad ?\tag{1}$$
In order to prove that this is not the case we make use of the underlying real structure. From $\bar z=x-iy$ it follows that $${\partial\bar z\over\partial x}=1,  \qquad{\partial\bar z\over\partial y}=-i\ ,\tag{2}$$
whereas on the right hand side we obtain, using the CR equations, that
$${\partial f\over\partial x}=u_x+iv_x,\qquad {\partial f\over\partial y}=u_y+iv_y=-v_x+iu_x\ .\tag{3}$$
If we assume $(1)$ then comparing the first equations of $(2)$ and $(3)$ leads to $u_x=1$, $v_x=0$, and comparing the second equations leads to $u_x=-1$, $v_x=0 \ $ – a contradiction.