Geometric Derivation of the D-Bar Operator $\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial }{\partial x} – i\frac{\partial }{\partial y})$

This picture

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from Visual Complex Analysis is all you need to derive the Cauchy-Riemann equations, i.e. from the picture we see $i \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$ should hold so we have

$$i \frac{\partial f}{\partial x} = i \frac{\partial (u+iv)}{\partial x} = \frac{\partial (u+iv)}{\partial y} \rightarrow C \ R \ Eq’s$$

Is there a similar picture-derivation of the operators

$$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial }{\partial x} – i\frac{\partial }{\partial y})$$

$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial }{\partial x} + i\frac{\partial }{\partial y})?$$

The fact the differential forms can be visualized in terms of sheets tells me there can be one, any ideas?

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Here is a trigonometric explanation that is intuitive for me.

Your initial basis vectors are $2$ vectors of length $1$, $x=(1,0)$ (parallel to $OX$ axis) and $y=(0,1)$ (parallel to $OY$ axis). This is what you have on your left hand side zoom picture.

If you turn them by $45$ degrees clockwise (NOT as shown on your right hand side zoom picture, in the opposite direction), you will have $z=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}(x+iy)$ and $\bar{z}=\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}(x-iy)$ vectors respectively. The corresponding differentials will be
$\frac{d}{dz} = \frac{1}{\sqrt{2}}\left(\frac{d}{dx}-i\frac{d}{dy}\right)$ and
$\frac{d}{d\bar{z}} = \frac{1}{\sqrt{2}}\left(\frac{d}{dx}+i\frac{d}{dy}\right)$ respectively. For this rotation, we use, $\theta=\frac{\pi}{4}$.

If you want to use $x\pm iy$ as a basis, you are stretching the basis defined in the previous paragraph by $\sqrt{2}$ and so you have to stretch the differentials by $\frac{1}{\sqrt{2}}$ to compensate, $$\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\left(\frac{d}{dx}\pm i\frac{d}{dy}\right) = \frac{1}{2}\left(\frac{d}{dx}\pm i\frac{d}{dy}\right) $$