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The Cauchy-Riemann condition states that an analytic function satisfies: \begin{split} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{split}

The converse of the statement requires additional conditions: the first partial derivatives of $u, v$ exist and are continuous.

I derived the Cauchy-Riemann condition based on the fact that partial derivatives of $f(z)$ with respect to $x, y$ $(z=x+iy)$ must be the same on the point $z$ where $f(z)$ is differentiable.

On the other hand, I found something very interesting that I cannot intuitively interpret. I was reading the Complex Variables and Applications (Brown and Churchill) 9th ed textbook [1]. On p.71-72, the author explains:

If $z=x+iy$, then $x=\frac{z+\overline z}{2}$ and $y=\frac{z-\overline z}{2i}$

By formally applying the chain rule in calculus to a function $F(x,y)$ of two real variables, we can derive the expression

$\frac{∂F}{∂\overline z}$ =$\frac{∂F}{∂x}$$\frac{∂x}{∂\overline z}$ + $\frac{∂F}{∂y}$$\frac{∂y}{∂\overline z}$ =$\frac{1}{2}(\frac{∂F}{∂x}+i\frac{∂F}{∂y})$

Now let's define the operator $\frac{∂}{∂\overline z} = \frac{1}{2}(\frac{∂}{∂x}+i\frac{∂}{∂y})$

By applying this operator to a analytic function $f(z) = u(x,y) + iv(x,y)$ which satisfies the Cauchy-Riemann conditions, then $\frac{∂f}{∂\overline z} = \frac{1}{2}[(u_x-v_y)+i(v_x+u_y)] = 0$

Thus derive the complex form $\frac{∂f}{∂\overline z} = 0$ of the Cauchy-Riemann equations.

I can follow the above proof step by step but I think there's much more to say about the result $\frac{∂f}{∂\overline z} = 0$. Why this result came from the Cauchy-Riemann condition? Is there any geometrical meaning more than just writing a list of formulas?

On the Wikipedia page - Cauchy Riemann equations(https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations), some comments are posted like this.

In this form, the Cauchy–Riemann equations can be interpreted as the statement that f is independent of the variable $\overline z$. As such, we can view analytic functions as true functions of one complex variable as opposed to complex functions of two real variables.

But it's quite confusing for me to understand and many questions remained. How can I view analytic functions(that satisfies the Cauchy-Riemann condition) as one complex variable and how such interpretation is directly related to the concept of differentiability?

Reference

James Brown, Ruel Churchill. Complex Variables and Applications 9thd ed; McGraw-Hill Education, 2013. ISBN 978-0073383170.

Arete
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    All complex valued functions are technically subsets of two variable functions $f(z,\bar{z})$ which are analytic, smooth, etc in the real sense. This is because we technically have a function of two variables $x$ and $y$, and we cannot map them to just one variable under a coordinate transformation. You also get these extended theorems: $$\int_{\partial \Omega}f:dz = 2i\iint\limits_\Omega \partial_{\bar{z}} f:dx\wedge dy$$ and $$f(z_0) = \frac{1}{2\pi i}\int_{\partial\Omega}\frac{f}{z-z_0}dz - \frac{1}{\pi}\iint\limits_{\Omega} \frac{\partial_{\bar{z}}f}{z-z_0}dx\wedge dy$$ – Ninad Munshi Sep 18 '20 at 18:59
  • This is also why complex differentiable functions seem to be so nice, it's because we put such a heavy restriction on the original space (functions that don't shear). Intuitively, this will measure the amount of "twisting" (according to some geometric metric) in the function space. This notion of extra terms measuring a twist of sorts carries over into more advanced math. – Ninad Munshi Sep 18 '20 at 19:03
  • there is simple geometrical interpretation: https://math.stackexchange.com/a/4439868/532409 – Quillo Apr 30 '22 at 14:29

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