Wirtinger derivative form of Cauchy–Riemann equations 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?

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: $$\frac{\partial}{\partial x}\log|z|
=\frac12\frac{\partial}{\partial x}\log(x^2+y^2)=\frac x{x^2+y^2}.$$
