TITLE: Why are analytic functions functions of $z$ and not of $\bar{z}$?
QUESTION [9 upvotes]: I was reading a note on complex analysis and was stuck on  one line:   

Cauchy-Riemann equations
Alternatively, the C-R equations can be written as
  $$\frac{\partial f}{\partial\bar z}=0,$$
  where $z=x+iy$ and $\bar z=x-iy$.
In some sense, analytic functions are truly functions of $z$, and not of $\bar z$.
  $$\begin{array}{c}
\frac\partial{\partial z} = \frac 1 2 \left(\frac\partial{\partial x}-i\frac\partial{\partial y}\right)
&\frac\partial{\partial \bar z}=\frac 1 2 \left(\frac\partial{\partial x}+i\frac\partial{\partial y}\right)
\end{array}$$

I do not understand what it means that "Analytic functions are truly functions of $z$ and not of $\bar{z}$." Can someone explain? Thanks in advance for your time.

REPLY [18 votes]: Editing slightly (thanks @Matt E): An alternative approach to give meaning to the sentence is this. A complex-valued real analytic function $f$ can be expanded in a power series in $x$ and $y$, hence, by substitution, in $z$ and $\bar z$. $f$ is complex analytic precisely when no $\bar z$'s appear in the latter expression. For example, $f(x,y)=(x^2-y^2)+2ixy = z^2$ is, of course, complex analytic, but $f(x,y)=(x^2+y^2)-2ixy = z\bar z +\frac12(\bar z^2-z^2)$ is not. Of course, this is equivalent to the calculation $\partial f/\partial\bar z \overset{?}{=} 0$.