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I asked this question: re expressing the Cauchy Riemann Equations

$$ \begin{split} \frac{\partial f}{\partial z} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial z}\\ &= \frac12 \left(\frac{\partial f}{\partial x} -i \frac{\partial f}{\partial y} \right) \end{split} $$

Where I asked why the second equation holds.

and this was the awnser given: To deduce the second equality it is sufficient to note that, since $z=x+iy$ (and $\bar{z}=x-iy$), then $$ x=\frac{1}{2}(z+\bar z)\quad y=-\frac{i}{2}(z-\bar z) $$ so $$ \frac{\partial x}{\partial z}=\frac{1}{2}\quad\frac{\partial y}{\partial z}=-\frac{i}{2} $$

I am still confused by one thing, to me it seems that the derivative of $\bar{z}$ does not exist, we can get both 1 and -1 at a same point approaching it either by the reals or the imarginaries.

So how would one get the derivative of: $z - \bar{z}$ with respect to $z$?

Thank you for the help!

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The point is that it takes place after you complexify, i.e., instead of $x,y\in\mathbb{R}$, you let $x,y\in\mathbb{C}$ and consider the change of coordinates from $x,y$ to $(z,\bar{z})=(x+iy,x-iy)$ as a purely algebraic manipulation (it has geometric interpretations too, but let's ignore that for now). Then $z,\bar{z}$ are independent coordinates, so $\dfrac{\partial\bar{z}}{\partial z}=0$ (note that it is $\partial$ not $\mathrm{d}$, so there isn't much risk of confusion).

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