Why can we view $z$ and $\bar z$ as independent variables in complex analysis?

I am quite confused about how to understand $\frac{\partial f}{\partial z}f(z,\bar z).$ Do $z$ and $\bar z$ in $f(z,\bar z)$ act the same way as $x$ and $y$ in $f(x,y)$?

If so, how can we prove this?

Solution 1:

Given an open subset $U\subset \mathbb C=\mathbb R^2$ and a $\mathcal C^\infty$-function $f:U\to \mathbb C$, one defines
$$ \frac{\partial f}{\partial z} = \frac{1}{2}\left(\frac{\partial f}{\partial x} -i\frac{\partial f}{\partial y}\right) \in \mathcal C^\infty (U) \; \text {and} \frac{\partial f}{\partial \bar z} = \frac{1}{2}\left(\frac{\partial f}{\partial x} +i\frac{\partial f}{\partial y}\right)\in \mathcal C^\infty (U)$$Notice carefully that I wrote $\mathbb C=\mathbb R^2$, an equality not an isomorphism: one has the same set but the notation $\mathbb C$ means we have endowed $\mathbb R^2$ with with its well-known field structure. Consequently one also writes $z=x+iy=(x,y)$.

This is essentially all there is to say. No mystery here: we have just defined two differential operators $ \frac{\partial f}{\partial z}, \frac{\partial f}{\partial \bar z} \in Der _{\mathbb C} (\mathcal C^\infty (U)) $.
There is an analogous punctual version $ \frac{\partial }{\partial z}\mid _{z_0}, \frac{\partial }{\partial \bar z}\mid _{z_0} \in Der _{\mathbb C} (\mathcal C^\infty _{z_0},\mathbb C) $ for functions defined only on a neighbourhood of a fixed point $z_o\in \mathbb C$
[$Der$ stands for derivation, a fancy algebraization of the good old Leibniz rule for taking the derivative of a product]

For example we have $\frac{\partial (x^2)}{\partial z}=x,\quad \frac{\partial (\sin xy +i e^ x)}{\partial \bar z}=\frac{1}{2}[y\cos xy+i(e^x+x\cos xy)]$

As I am sure you know, a $\mathcal C^{\infty} $ function $f$ is holomorphic iff $\frac{\partial f}{\partial \bar z}=0 $ and in that case $f'(z)=\frac{\partial f}{\partial x}(z)$.
For example if $f(z)=z^2=x^2-y^2+2ixy$ then $f'(z)=2z=2x+2iy=\frac{\partial (x^2-y^2+2ixy)}{\partial x}$

And what about $\frac{\partial }{\partial z}f(z,\bar z)$ ? Forget about that notation : it makes absolutely no sense if $z$ is not real, because already $f(z,\bar z)$ is absolutely not defined !
[Actually, there are contorsions which define $f(z,\bar z)$ for real-analytic functions like polynomials, but they are artificial, hide the utmost simplicity of the Wirtinger calculus (that's the name of the guy who introduced the partials $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \bar z}$) and thus should be avoided]
Edit
In the same vein, $z$ and $\bar z$ are not at all independent. Quite the contrary: $\bar z$ is completely determined by $z $ !
What people mean when they use that ridiculous phrase is probably that $\frac{\partial z}{\partial \bar z}=0$, but then they should say just that and not introduce this absurd terminology of "independent variables".
[I have only noticed now that this was your actual question! Sorry for that.]

Solution 2:

Consider a function $f:\mathbb C\rightarrow \mathbb C$, use the identification $\mathbb C\simeq\mathbb R^2$ to write $$ f(x+iy)=(u(x,y),v(x,y)), $$ and assume it is differentiable w.r.t to $x$ and $y$. By definition, $$ \frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} -i\frac{\partial}{\partial y}\right),\qquad \frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} +i\frac{\partial}{\partial y}\right), $$ which means that, if $z_0=x_0+iy_0$, $$ \frac{\partial}{\partial z}f(z_0)=\frac{1}{2}\left(\frac{\partial}{\partial x}u(x_0,y_0)+\frac{\partial}{\partial y}v(x_0,y_0),\frac{\partial}{\partial x}v(x_0,y_0)-\frac{\partial}{\partial y}u(x_0,y_0)\right) $$ and $$ \frac{\partial}{\partial \bar z}f(z_0)=\frac{1}{2}\left(\frac{\partial}{\partial x}u(x_0,y_0)-\frac{\partial}{\partial y}v(x_0,y_0),\frac{\partial}{\partial x}v(x_0,y_0)+\frac{\partial}{\partial y}u(x_0,y_0)\right). $$ From this, you may check that $$ \frac{\partial}{\partial z}\bar z=0,\qquad \frac{\partial}{\partial \bar z}z=0, $$ which shows the "independence" between $z$ and $\bar z$ to be similar than the one between $x$ and $y$ you were looking for.

These derivatives are pretty convenient to manipulate, and note that if $f$ is holomorphic, namely satisfies the Cauchy-Riemann equations $$ \frac{\partial}{\partial x}u(x_0,y_0)=\frac{\partial}{\partial y}v(x_0,y_0),\quad \frac{\partial}{\partial x}v(x_0,y_0)=-\frac{\partial}{\partial y}u(x_0,y_0), $$ then $$ \frac{\partial}{\partial z}f(z_0)=\left(\frac{\partial}{\partial x}u(x_0,y_0),\frac{\partial}{\partial y}v(x_0,y_0)\right),\qquad \frac{\partial}{\partial \bar z}f(z_0)=0. $$

Solution 3:

All this makes perfect sense; you just have to use the right definitions:

Let $f: \mathbb{C}\simeq \mathbb{R}^2\to \mathbb{C}$ be real analytic, that is, if you think of it as being defined on $\mathbb{R}^2$ it has a power series expansion in $(x,y)$ at every point in $\mathbb{R}^2$ (note that this is weaker than being holomorphic). Then $f$ can be extended to a holomorphic function $D\to\mathbb{C}$, where $D\subset \mathbb{C}^2$ is a neighborhood of $\mathbb{R}^2 \subset \mathbb{C^2}$, so you can write $f(x,y)$ where $x,y$ are now complex! Now consider the following change of variables: $$(z,\bar z)=(x+iy,x-iy)$$ Here we regard $\bar z$ as just a symbol that is independent of $z$. Then it makes sense to write $f(z,\bar z)$; $\partial_z$, $\partial_{\bar z}$ are the Wirtinger derivatives, and $\partial_{\bar z}f=0$ (that is, $f$ is actually independent of $\bar z$) is equivalent to $f$ being holomorphic.