Q: Simple question on complex derivative For any complex function $f(z)$, let $z=x+iy$, and we can write $f(z)=g(x,y)=u(x,y)+iv(x,y)$. Hence, $x=\frac{1}{2}(z+\bar{z})$ and $y=\frac{1}{2i}(z-\bar{z})$. Then $\frac{\partial f}{\partial z} = \frac{1}{2}(\frac{\partial g}{\partial x} - i \frac{\partial g}{\partial x})$. Is the last step true because that we view $g(x,y)$ as a multivariate function so we can use chain rule about $x$ and $y$ with respect to z? But is this chain rule true for complex function, even if $x$ and $y$ is real? Basically, can anyone explain why $\frac{\partial f}{\partial z} = \frac{1}{2}(\frac{\partial g}{\partial x} - i \frac{\partial g}{\partial x})$ is true for complex function?

A: To my understanding, 
$$\frac{\partial g}{\partial z} = \frac{1}{2}\left(\frac{\partial g}{\partial x} - i \frac{\partial g}{\partial y}\right) \tag{1}$$ is a definition of the left-hand side. So is 
$$\frac{\partial g}{\partial \bar z} = \frac{1}{2}\left(\frac{\partial g}{\partial x} + i \frac{\partial g}{\partial y}\right) \tag{2}$$
The reason to introduce the above: we need differential operators $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \bar z}$ which can separate holomorphic things from those that aren't. Specifically, we need
$$\frac{\partial }{\partial z} z =1 ,\quad \frac{\partial }{\partial \bar z} \bar z =1$$
and 
$$\frac{\partial }{\partial z}\bar z  =0 ,\quad \frac{\partial }{\partial \bar z}   z =0$$
Linear algebra manipulations lead from here to (1) and (2).
