Derivative of the complex norm as commonly used in physics On the one hand, I read that the derivative of the complex conjugate $C[z]=\overline{z}$ is not differentiable anywhere (for instance see here). (see 1, below)
On the other hand, I see in physics taking the derivative of a complex scalar field to obtain the equation of motion using the Euler-Lagrange method (for instance see enter link description here (see 2, below)
So which is it, can we or can we not take the derivative?


*

*For case 1, the reference states that a complex function is differentiable if and only if it satisfies the Cauchy-Riemann equations:

$$
f[z]=f[x+iy]=u[x,y]+iv[x,y]
$$
Then f is differentiable if
$$
\frac{\partial u}{\partial x} =\frac{\partial v}{\partial y} \\ 
\frac{\partial u}{\partial y} =-\frac{\partial v}{\partial x}
$$
Then for the complex conjugate $C[x+iy]=x-iy$ then $\partial u/\partial x =1$ and $\partial v/\partial y=-1$. Consequently $C[z]=\overline{z}$ is not differentiable anywhere in the complex plane.


*For case 2, the physics paper defines the Lagrangian of a complex scalar free field as follows:

$$
\mathcal{L}=(\partial \phi^*)(\partial \phi)
$$
Then they claim that
$$
\frac{\partial \mathcal{L}}{\partial (\partial \phi)}=\partial \phi^*\\
\frac{\partial \mathcal{L}}{\partial (\partial \phi^*)}=\partial \phi
$$

To obtain these results I assume they apply the chain rule
$$
\frac{\partial }{\partial (\partial \phi)}(\partial \phi^* \partial \phi)=\partial \phi\frac{\partial }{\partial (\partial \phi)}(\partial \phi^* )+\partial \phi^* \frac{\partial }{\partial (\partial \phi)}(\partial \phi)
$$
Is the following term not an 'illegal' derivative of a complex conjugate function?
$$
\partial \phi\frac{\partial }{\partial (\partial \phi)}(\partial \phi^* )
$$
Why are they allowed to pose it equal to 0?
 A: I've already recommended a discussion here, but it may be worth rewriting to tweak its emphasis. I'll compare three contexts in which one would want to define "derivatives"; in each case, I'll consider a function $f(x)$, rather than switching from the label $x$ to $z$ for the complex case.

*

*A function $f:\,\Bbb R\mapsto\Bbb R$ differentiable at $x$ satisfies $f(x+h)\in f(x)+[(Df)(x)]h+o(h)$, with $Df(x)\in\Bbb R$ denoting the derivative of $f$ at $x$.

*For an integer $n\ge2$, A function $f:\,\Bbb R^n\mapsto\Bbb R^n$ differentiable at $x$ satisfies what looks like the same condition, only now $Df(x)\in\Bbb R^{n\times n}$.

*But if $S$ is a number system $n$-dimensional over $\Bbb R$, a function $f:\,S\mapsto S$ might satisfy a stronger condition, that we can take $Df(x)\in S$ where, as in the original example, the $O(h)$ term multiplies a number, in this case one in $S$, by another such number, $h$. The Cauchy-Riemann equations characterize which functions $f:\,\Bbb C\mapsto\Bbb C$ succumb to this treatment.

The trick here is respecting the multiplication defined on $\Bbb C$. If we denote the real and imaginary parts of a compex number $w$ as $w_0$ and $w_1$, we cannot in general change$$f_j(x+h)\in f_j(x)+\sum_{k=0}^1[(Df)(x)]_{jk}h_k+o(h)$$to $f_j(x+h)\in f_j(x)+(yh)_j+o(h)$ for some $y\in\Bbb C$. But the general case succumbs to another treatment. Since any $w\in\Bbb C$ satisfies $w_0=(w+w^\ast)/2,\,w_1=(w-w^\ast)/(2i)$,$$\begin{align}f_j(x+h)- f_j(x)&\in[(Df)(x)]_{j0}(h+h^\ast)/2+[(Df)(x)]_{j1}(h-h^\ast)/2i+o(h)\\&=\frac12\left\{[(Df)(x)]_{j0}-i[(Df)(x)]_{j1}\right\}h\\&+\frac12\left\{[(Df)(x)]_{j0}+i[(Df)(x)]_{j1}\right\}h^\ast+o(h).\end{align}$$So now, instead of only having a rate of change against small $h$, we also have a separate one against $h^\ast$. And although $h,\,h^\ast$ are "dependent" in the sense either value determines the other, these two rates of change are uniquely defined. The abbreviation $x:=\partial\phi$ lets us work with the given example from physics:$$\mathcal{L}(x)=x^\ast x\implies\mathcal{L}(x+h)-\mathcal{L}(x)=(x^\ast+h^\ast)(x+h)-x^\ast x\in x^\ast h+xh^\ast+o(h).$$
