Why is $\frac{\partial \bar f }{\partial z} = 0$ for an analytic function $f$?

Why is $$\frac{\partial \bar f }{\partial z} = 0$$ for an analytic function $$f$$?

I understand that, for an analytic function $$f, \frac{\partial f }{\partial \bar z} = 0$$, but I do not get why the above is true. How can one justify "flipping" the complex conjugate sign (and also without actually defining $$\frac{\partial \bar f }{\partial z}$$ to be $$\overline{\frac{\partial f }{\partial \bar z}}$$)?

I get that $$\frac{\partial \bar f }{\partial z} = \frac 12 (\frac{\partial \bar f }{\partial x} - i \frac{\partial \bar f }{\partial y})$$, but I am not sure how this is related to Cauchy-Riemann equations.

• It's the Cauchy-Riemann equations. – Lord Shark the Unknown Nov 23 '18 at 4:38
• @LordSharktheUnknown could you elaborate? – Cute Brownie Nov 23 '18 at 4:54

So first let's write the CR equations with $$f = u+iv$$. $$u_x = v_y$$ $$-u_y = v_x$$ Then, as you noted in your last line, $$\frac{\partial \bar{f}}{\partial z} \sim \frac{\partial \bar{f}}{\partial x} - i\frac{\partial \bar{f}}{\partial y}$$ dropping the constant factor because we want to show this is $$0$$ anyway. Substituting for our functions $$u$$ and $$v$$, we have $$\begin{gather*} \frac{\partial \bar{f}}{\partial z} = \frac{\partial}{\partial x}\Big[u-iv\Big]-i\frac{\partial}{\partial y}\Big[u-iv\Big] \\ = u_x-iv_x-i\big(u_y-iv_y\big)\\ = u_x-v_y - i(v_x + u_y) \end{gather*}$$ This expression is $$0$$ by CR equations.
Just write $$\overline {f}=u-iv$$ where $$u$$ and $$v$$ are the real and imaginary parts of $$f$$. Then $$2\frac {\partial \overline {f}} {\partial z}$$ becomes $$(u_x-v_y)-i(u_y+v_x)$$ after some simplification. Both the terms are $$0$$ by C-R equations.