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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.

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

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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.

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