If $f$ is holomorphic, what is the meaning/intuition behind $f_z=f'(z)$ and $f_{\overline z}=0$?

Question

If $f$ is holomorphic then we know the derivative of $f$ with respect to $z$ is defined, i.e., $f'(z)$ exists. But $\overline{z}$ is a different variable, so if we take the derivative of $f$ with respect to a different variable, isn't it obvious the derivative will be $0$?

Is there something I'm failing to understand about if $f$ is holomorphic, then $f_z=f'(z)$ and $f_{\overline z}=0$?

Answer 1

1. Case $x,y,u(x,y),v(x,y) \in \mathbb{R}$ are all real-valued (as opposed to complex-valued). (This is the normal case.) The Dolbeault operators are defined as $$ \partial ~:=~\frac{1}{2}\left( \frac{\partial}{\partial x} -i\frac{\partial}{\partial y} \right) \quad\text{and}\quad \bar{\partial} ~:=~\frac{1}{2}\left( \frac{\partial}{\partial x} +i\frac{\partial}{\partial y} \right).\tag{A}$$ It is easy to see that the Cauchy-Riemann (CR) equations are satisfied iff $f\equiv u+iv$ satisfies $$\bar{\partial}f~=~0.\tag{B}$$ In the affirmative case, the function $f$ is complex differentiable/holomorphic, and $$ \frac{d f}{dz}~=~\frac{\partial f}{\partial x} ~=~\frac{1}{i} \frac{\partial f}{\partial y}~=~\partial f \tag{C}$$ holds. Note that $$ z~:=~x+iy \quad\text{and}\quad\bar{z}~:=~x-iy \tag{D}$$ are not independent variables.

2. Case $x,y,u(x,y),v(x,y) \in \mathbb{C}$ are all complex-valued (as opposed to real-valued). (This is the unusual case.) Define independent complex variables $z$ and $\bar{z}$ via eq. (D). Note that the bar in $\bar{z}$ no longer denotes complex conjugation. The inverse transformation reads $$ x~=~\frac{z+\bar{z}}{2}\quad\text{and}\quad y~=~\frac{z-\bar{z}}{2i}. \tag{E}$$ The chain rule yields $$ \frac{\partial}{\partial z} ~\stackrel{(E)}{=}~\frac{1}{2}\left( \frac{\partial}{\partial x} -i\frac{\partial}{\partial y} \right) \quad\text{and}\quad \frac{\partial}{\partial \bar{z}} ~\stackrel{(E)}{=}~\frac{1}{2}\left( \frac{\partial}{\partial x} +i\frac{\partial}{\partial y} \right),\tag{F}$$ which resembles eq. (A). For this reasons we often infuse some of the notation from case 2 into case 1 whenever no confusion can arise. See also this related Phys.SE post.