# Why is $\frac{df}{d\bar z}=0$ necessary if $f$ is an analytic function?

For an analytic function $f(z)$ with $z = x+iy$ one can derive the Cauchy Riemann conditions by using the interesting statement $$\frac{df}{d\bar z} = 0$$ I know how to derive the CR conditions from there on but what I don't really understand is why every $f$ must be independent of $\bar z$.
I don't feel it is enough to say that $f$ only depends on $z$ and that is why the statement must be true.
Since $z$ and $\bar z$ themselves both depend on $x$ and $y$ I wonder if there is deeper notion behind it.
My vague idea is that it maybe results from $x$ and $y$ being linearly independent but I can't really finish that thought.

• The key idea of "analyticity" is that $f$ is a pure function of $z$, rather than $z$ and $\overline{z}$ (in which case, we would really be thinking of $f$ as a function of $x$ and $y$ separately). – user296602 Feb 13 '18 at 18:32
• Oh, that makes sense. Thanks! – P-A Feb 13 '18 at 18:39

Cauchy-Riemann conditions: $\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$, then with $\dfrac{\partial f}{\partial x}=\dfrac{\partial u}{\partial x}+i\dfrac{\partial v}{\partial x}$ and $\dfrac{\partial f}{\partial y}=\dfrac{\partial u}{\partial y}+i\dfrac{\partial v}{\partial y}=-\dfrac{\partial v}{\partial x}+i\dfrac{\partial u}{\partial x}$, then \begin{align*} \dfrac{\partial f}{\partial\overline{z}}=\dfrac{1}{2}\left(\dfrac{\partial f}{\partial x}-\dfrac{1}{i}\dfrac{\partial f}{\partial y}\right)=0. \end{align*}