partial derivative with respect to $\overline{z}$ In my text on complex analysis, they give the definition of $\frac{\partial f}{\partial \overline{z}}$ for suitable $f : \mathbb{C} \rightarrow \mathbb{C}$. However, I do not understand how to make formal the fact that if you write $f$ 'in terms of' $z$ and $\overline{z}$, $\frac{\partial f}{\partial \overline{z}}$ is just the partial derivative with respect to $\overline{z}$ of $f$, as if you substituted $x=z$ and $y=\overline{z}$ and then took the partial with respect to $y$. The book uses this without proof, I would like to understand why this works. Thanks in advance.
 A: The reason your book uses the expressions $\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \overline{z}}$ without proof is that they are not actual partial derivatives, they are only derivatives in a symbolic sense. They are formal expressions that can help in understanding and communicating the behavior of a complex function. 
If we write $x = \frac12(z + \overline{z})$ and $y = \frac{1}{2i}(z - \overline{z})$, thus expressing $x$ and $y$ as functions of $z$ and $\overline{z}$, and think of $f(z) = f(x,y)$ as a function of two real variables, then by supposing the rules of calculus apply we obtain
$$
\frac{\partial f}{\partial z} = \frac 12 \left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right)
$$
and
$$
\frac{\partial f}{\partial \overline{z}} = \frac 12 \left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right).
$$
If $\frac{\partial f}{\partial \overline{z}} = 0$ for a particular function, then this is equivalent with saying that $f$ satisfies the Cauchy-Riemann equations. So in a formal sense we could say that an analytic function is a function of $z$ alone and independent of $\overline{z}$, thinking of them independent variables. (Ahlfors)
