# On understanding the Cauchy-Riemann equations

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!

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).