Ytukyg

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.

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,
$$
frac{partial bar{f}}{partial z} sim frac{partial bar{f}}{partial x} - ifrac{partial bar{f}}{partial y}
$$
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-ivBig]-ifrac{partial}{partial y}Big[u-ivBig] \
= u_x-iv_x-ibig(u_y-iv_ybig)\
= 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 $2frac {partial overline {f}} {partial z}$ becomes $(u_x-v_y)-i(u_y+v_x)$ after some simplification. Both the terms are $0$ by C-R equations.