How to understand the notation $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ (Wirtinger derivatives)? The Wirtinger derivatives are defined as $$\frac{\partial}{\partial z}:=\frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right)\quad \quad\frac{\partial}{\partial \bar{z}}:=\frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$$
I wonder why we use the notation partial derivative $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ to represent Wirtinger derivatives. Does this suggest Wirtinger derivatives are a kind of partial derivative with respect to $z$ or $\bar{z}$ ?
I am confused by this fact: $z$ and $\bar{z}$ are not independent. Here by not independent I mean $\bar{z}$ is a function of $z$. This leads to several following problems.

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*Let $f$ be the function $f(z)=\bar{z}$. As a function of $z$, the partial derivative of $f$ with respect to $z$ should not equal to 0. However, according the definition of Wirtinger derivatives, $\frac{\partial f}{\partial z} = 0$.

*Let $f:\mathbb{C} \to \mathbb{C}$ be a function. Can we write $f(z) = g(z,\bar{z})$ for some function $g:\mathbb{C}^2 \to \mathbb{C}$ and regard $\frac{\partial f}{\partial z}$ as $\frac{\partial g}{\partial z}$, where $\frac{\partial g}{\partial z}$ means partial derivative instead of Wirtinger derivative ? If so, when we compute partial derivative of $g$ respect to $z$, we should fix the second variable $\bar{z}$. But fixing $\bar{z}$ also means fixing $z$, then how does partial derivative make sence?

*By the definition of Wirtinger derivatives, it's easy to get $\frac{\partial \bar{z}^n}{\partial \bar{z}} = n\bar{z}^{n-1}$, similar to partial derivative. For arbitrary $f:\mathbb{C} \to \mathbb{C}$, can we get a right answer like this, computing Wirtinger derivative as partial derivative ? If so, how to prove it?

I believe the notation $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ suggest Wirtinger derivatives are a kind of partial derivative, but I can't solve the problems above. Could you please give me a rigorous answer? Thanks for your help.
 A: You and I both know that the tooth fairy does not exist buttttttt let's suppose that it did and let's see if we can deduce anything useful about the world from that.
This, essentially, summarizes the way in which we justify creating these so-called Wirtinger derivatives and giving them the symbols that we do. It's one of those things that you come up with when you're randomly messing around with a bunch of formulas without justifying what you're doing but you're also born with the intellect of Riemann so you can kind of create something useful out of it.
The idea here is to take a holomorphic function $f:U \to \mathbb{C}$ defined on an open subset of $\mathbb{C}$. This is actually a two-variable function $(x,y) \mapsto f(x,y)$. Now, $x = \frac{1}{2}(z+\overline{z})$ and $y = \frac{1}{2i}(z-\overline{z})$, whenever $z = x+iy$. Now, here's the main idea that is being made use of in the following (heuristic) derivation:
Idea: $z$ and $\overline{z}$ should be regarded as independent variables.
So, they're not independent, as you correctly pointed out. There is no tooth fairy. Buttttttt, let's assume that they were independent. Then, $x$ and $y$ could also be written individually as functions of $z$ and $\overline{z}$. So, essentially, we have the following situation:
$$(z,\overline{z}) \mapsto (x(z,\overline{z}),y(z,\overline{z})) \mapsto f(x,y)$$
and this means that $f$ can be written as a function of $z$ and $\overline{z}$, as you have also asked about in your second question, and we can now differentiate with respect to each of these variables using the chain rule. So:
$$\partial_z f = \partial_x f \cdot \partial_z x + \partial_y f \cdot \partial_{z} y = \frac{1}{2} \partial_x f - \frac{1}{2}i \partial_y f = \frac{1}{2} (\partial_x f -i\partial_y f)$$
All I've done is to treat $z$ and $\overline{z}$ as independent variables (that is the entire thesis of the calculus of Wirtinger  derivatives). Similarly:
$$\partial_{\overline{z}} f = \partial_x f \cdot \partial_{\overline{z}} x + \partial_y f \cdot \partial_{\overline{z}}y = \frac{1}{2} \partial_x f + \frac{1}{2} i \partial_y f = \frac{1}{2}(\partial_x f + i\partial_y f) $$
Can you see how the Wirtinger derivatives arise naturally in the context above? But now, the main question that you asked is still floating in the air:
Does this suggest Wirtinger derivatives are a kind of partial derivative with respect to $z$ and $\overline{z}$?
If we can address this, then all your other questions can be addressed easily. Judging by all of the information above and the heuristics of the derivation above, the answer is a resounding No. You see, the Wirtinger derivatives don't represent partial differentiation with respect to $z$ or $\overline{z}$. As you've pointed out repeatedly, they are not independent.
The derivation above, done in the most naive manner, just suggests that the operators $\frac{1}{2}(\partial_x+i\partial_y)$ and $\frac{1}{2}(\partial_x -i\partial_y)$ are important. That's what's being suggested over here (and in fact, the Cauchy-Riemann Equations also do suggest this, let me know if you want further explanation on that). But now, when you have operators of interest, then you want to give them names.
But what names should we pick? How should we encode the fact that these operators are "somewhat" behaving like some sort of "partial differentiation with respect to $z$ or $\overline{z}$"? Well, that's where you use the symbols $\partial_z$ and $\partial_{\overline{z}}$ as homage to the intuitive derivation above.
So, you see, the Wirtinger derivatives don't represent partial differentiation. They represent homage to the heuristic above that aims to treat the idea of partial differentiation with respect to $z$ or $\overline{z}$ seriously. Now, what does this actually mean for your questions?

*

*Your first two questions are certainly addressed by the considerations above. This isn't partial differentiation. So, if I give you a function $f(x,y)$ and ask you to compute $\partial_z f$, what you'll do is to use the definition of the Wirtinger derivative to compute this derivative.

But, in fact, that is tantamount to using the heuristic that $z$ and $\overline{z}$ are to be treated as independent variables. Indeed, in your first question, you showed that $\partial_z(\overline{z}) = 0$ and it's also not too hard to show that $\partial_{\overline{z}}(z) = 0$ as well. If we "think of" these as referring to some kind of differentiation, this just shows that $z$ and $\overline{z}$ are "independent".


*Now, your third question is much more interesting. But the way to summarize this is the following; what you really want is a theorem that shows you that these two operators actually do behave like some kind of differentiation. Indeed, such a theorem does exist.

Theorem:
$\partial_z, \partial_{\overline{z}}$ are $\mathbb{C}$-linear mappings for which the product and quotient rule hold.
Of course, you should define these operators on the appropriate vector spaces. This is actually very easy to prove and I'll leave the proof of it to you. You can certainly find results of this form online in multiple documents and these are the results that you typically use to find Wirtinger derivatives for functions. They turn out to correspond very well with the usual rules for partial differentiation.
