What is the derivative of the modulus of a complex function? For $|f(z)| = |z|$, where $z = x + iy$, $x$, $y$ real, it is known that the modulus $|z|$ is complex differentiable only at $z = 0$, i.e. $x = y = 0$.
My question concerns the differentiability of the modulus of a general complex function $|f(z)|^2 = |U(z) + iV(z)|^2 = U^2 + V^2$.
I have found no reference that deals with this question. My reasoning is that this differentiability should be at $U = V = 0$, not at $x = y = 0$
Related questions: (1) what is the expression of this derivative? 
(2) Is this derivative equal to zero? I would say yes, it is.
 A: The reference that suggests that $|z|$ is differentiable at $z=0$ is incorrect.  Note that if $f(z)$ is differentiable, then $$f'(z)=\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$$
Taking $f(z)=|z|$ and $z=0$, we see that $\displaystyle \lim_{\Delta z\to 0}\frac{|\Delta z|-0}{\Delta z}$ fails to exist.
Similarly, if $f(z)=u(x,y)+iv(x,y)$, and $g(z)=|f(z)|^2=u^2(x,y)+v^2(x,y)$, then $g(z)$ is purely real.  The only purely real function that is complex differentiable in an open neighborhood of a point is a function that is constant.  So, $g$ is differentiable in a neighborhood of $z$ only if $f$ is constant there.
To show this, we appeal to the Cauchy-Riemann equations.  Note that if $h(z)=\phi(x,y)$, where $\phi$ is a purely real-valued function, then the Cauchy-Riemann equations reveal 
$$\frac{\partial \phi(x,y)}{\partial x}=\frac{\partial \phi(x,y)}{\partial y}=0\implies \phi(x,y)\,\,\text{is a constant}$$

Now, it is possible that $|f|^2$ might be differentiable at an isolated point $z$, but not in an open neighborhood of that point.  Note that if $f$ is analytic in the open domain $O$, and $z\in O$ and $z+\Delta z\in O$, then 
$$\begin{align}
\frac{|f(z+\Delta z)|^2-|f(z)|^2}{\Delta z}&=\frac{|f(z)+f'(z)\Delta z+O\left((\Delta z)^2\right)|^2-|f(z)|^2}{\Delta z}\\\\
&=\frac{2\text{Re}\left(\overline{f(z)}f'(z)\Delta z\right)}{\Delta z}+O\left(\Delta z\right)
\end{align}$$
Hence, if the limit $\lim_{\Delta z\to 0}\frac{2\text{Re}\left(\overline{f(z)}f'(z)\Delta z\right)}{\Delta z}$ exists, then $|f|^2$ is differentiable at $z$.  Again, $|f|^2$ cannot be analytic at any point unless $f$ is a constant.
A: I hope that following will help clarify some things.
Let us define $$g\left( z \right) = {\left| z \right|^2}$$
We show that this function is complex differetiable at the origin.
Indeed,
$$g'\left( 0 \right) = \mathop {\lim }\limits_{z \to 0} {{g\left( z \right) - g\left( 0 \right)} \over {z - 0}} = \mathop {\lim }\limits_{z \to 0} {{{{\left| z \right|}^2} - 0} \over {z - 0}} = \mathop {\lim }\limits_{z \to 0} {{z \cdot \overline z } \over z} = \mathop {\lim }\limits_{z \to 0} \left( {\overline z } \right) = 0$$
Thus $g\left( z \right)$ is complex differetiable at the origin and its derivative there is zero.
Notice that $g\left( z \right)$ is not constant.
An important remark is that a function can be complex differentiable at a point and still not analytic/holomorphic at that point. The above $g\left( z \right)$ is an example of such a function.
We say that $g\left( z \right)$ is holomorphic at the point ${z_0}$ if and only if it is complex differentiable on some neighborhood of ${z_0}$.
Our function $g\left( z \right)$ satifies the cauchy-riemann equations only at the origin and thus cannot be holomorphic there. 
(The Cauchy-Riemann equation are a necessary but not a sufficient condition for complex differentiability)
