Derivatives by complex number and conjugate

Solution 1:

Let me expand Daniel Fischer's hint.

$z = x + iy \rightarrow \bar{z} = x - iy$ which gives $x = \frac{1}{2}(z + \bar{z})$ and $y = \frac{1}{2i}(z - \bar{z})$. So we may consider the function $f(x,y)$ as a function of $z$ and $\bar{z}$.

Differentiating the relations we shall get $\frac{\partial{x}}{\partial{z}} = \frac{1}{2}$, $\frac{\partial{y}}{\partial{z}} = \frac{1}{2i}$, $\frac{\partial{x}}{\partial{\bar{z}}} = \frac{1}{2}$ and $\frac{\partial{y}}{\partial{\bar{z}}} =\frac{-1}{2i}$

Now use chain rule

$$\frac{\partial{f}}{\partial{z}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{z}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{z}} = \frac{1}{2}\left(\frac{\partial{f}}{\partial{x}} - i\frac{\partial{f}}{\partial{y}}\right)$$

Similarly applying the chain rule for $\bar{z}$ you shall get another result.

Here we shall assume that all the rules of calculus applied here is applicable.

Solution 2:

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 $2.3. \textit{Conformal Mapping.}$