If $f$ is holomorphic then it derivative equals the first Wirtinger derivative?

Question

Sorry if this question is so dumb but, strangely, I cannot find some reference to it. I have the Wirtinger derivatives defined by

$$\partial_W f:=\frac12(\partial_x f-i\partial_y f),\quad\overline\partial_W f:=\frac12(\partial_x f+i\partial_y f)$$

And the complex differential of $f:=u+iv$ for some $u,v\in C^1(U,\Bbb R)$ for some $U\subset\Bbb C$ is defined by $df:=(u_x+iv_x)dx+(u_y+iv_y)dy$.

Then its easy to see that when $f$ is holomorphic then $\overline\partial_W f=0$, but also we have that $f' dz=\partial_W f dz+\overline\partial_W fd\overline z$ for $d\overline z:=dx-idy$.

This means that $f'=\partial_W f$, right?

Answer 1

Yes, this is the purpose one having these two differential operators. They are convenient because

1. Both are defined for arbitrary $\mathbb{R}$-differentiable functions.
2. One expresses the Cauchy-Riemann equations, its vanishing characterizes holomorphic functions.
3. The other expresses the complex derivative whenever it exists.