Wirtinger derivative form of Cauchy–Riemann equations

Question

I'm trying to understand the Cauchy-Riemann equations using the traditional $u, v$ form and the Wirtinger derivative form.

Taking $\ln|z|$ as an example function, for the normal $u, v$ form I have: $$\begin{align}u(x,y) &= \ln|x + iy|\\ v(x,y) &= 0\end{align}$$ so the Cauchy-Riemann equations are not satisfied: $$\frac{\partial}{\partial x} u(x,y) = \frac{1}{x + iy} \neq v \frac{\partial}{\partial y} = 0$$ $$\frac{\partial}{\partial y} u(x,y) = \frac{i}{x + iy} \neq -v \frac{\partial}{\partial x} = 0$$

So far so good, I didn't expect them to be. But there's another form for the Cauchy-Riemann equations using Wirtinger derivatives:

$$\frac{\partial}{\partial \overline{z}} f(z) = 0$$

Doing it this way I get

$$\begin{align} \frac{\partial}{\partial \overline{z}} \ln|z| &= \\ &= \frac{1}{2} (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}) \ln|x + iy| \\ &= \frac{1}{2} (\frac{1}{x + iy} + i \frac{i}{x + iy}) \\ &= \frac{1}{2} (\frac{1}{z} - \frac{1}{z}) \\ &= 0\end{align}$$

So using the Wirtinger derivative form it would seem that $\ln|z|$ is holomorphic? I don't think that's right; I thought real valued functions should only be holomorphic if they're constant. What am I doing wrong?

Answer 1

As commented, you got some of the partial derivatives wrong.

A trick that's often useful: $$\log|z|=\frac12\log|z|^2=\frac12\log(x^2+y^2).$$ Useful because we know how to differentiate polynomials: $$\frac{\partial}{\partial x}\log|z| =\frac12\frac{\partial}{\partial x}\log(x^2+y^2)=\frac x{x^2+y^2}.$$

Answer 2

You have already got an answer how to do it correctly. But I think that one thing is missing: What is wrong with your way?

When you take the derivative of $\ln |x+iy|$ using the chain rule, you actually assume that $\ln |z|$ is differentiable: $$ \frac{\partial}{\partial x} \ln |z| = \underbrace{\frac{d\ln |z|}{dz}}_{\text{invalid!}} \frac{\partial z}{\partial x} $$

Using this error we could "show" that any real-differentiable function is complex-differentiable: $$ \frac{\partial}{\partial\bar z} f(z) = \frac{1}{2} (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}) f(x+iy) = \frac{1}{2} (f'(x+iy) \cdot 1 + i f'(x+iy) \cdot i) = 0. $$