TITLE: Derivatives by complex number and conjugate
QUESTION [15 upvotes]: Suppose we write a complex number $z=x+iy$. Then we have $\overline{z}=x-iy$, and therefore $x=\dfrac12(z+\overline{z})$, $y=-\dfrac12i(z-\overline{z})$. 

If the rules of calculus were applicable, we would obtain $$\frac{\partial{f}}{\partial{z}}=\frac12\left(\frac{\partial{f}}{\partial{x}}-i\frac{\partial{f}}{\partial{y}}\right), \frac{\partial{f}}{\partial{\overline{z}}}=\frac12\left(\frac{\partial{f}}{\partial{x}}+i\frac{\partial{f}}{\partial{y}}\right)$$

How does this follow from the rules of calculus?

REPLY [10 votes]: This idea is only $\textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $\bar{z}$ by the definition would automatically change, since $\bar{z}=\frac{|z|^2}{z}$. The variables $z$ and $\bar{z}$ are not independent, and we can not perform the $\textit{partial derivations}$, by keeping one as constant. So technically $\frac{\partial x}{\partial z} \neq \frac{1}{2}$ and so as the other such partial derivatives. One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$\frac{\partial f}{\partial z}:= \frac{1}{2}\Bigg(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\Bigg)\text{ and } \frac{\partial f}{\partial \bar{z}}:= \frac{1}{2}\Bigg(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\Bigg).$$ It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $\frac{\partial f}{\partial \bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $\textit{Complex Analysis}$ by $\textit{Lars V. Ahlfors}$, section $2.3. \textit{Conformal Mapping.}$