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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.

Dando18
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Hass
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    Your question is unclear. Did you mean to write $f(z) = \lvert z \rvert$? If so, the absolue value on $ℂ$ is nowhere complex differentiable, and especially not in zero where it is not even partially differentiable. Also $\lvert U + \mathrm i V \rvert ≠ U^2 + V^2$; you’d have to square the left side for the equation to hold. – k.stm Aug 25 '17 at 13:21
  • Yes. I mean the modulus of a complex function. This reference says that |z| is differentiable only at z = 0: – Hass Aug 25 '17 at 13:27
  • Do you know the cauchy reimann equations? You can determine where any complex function is differentiable. – Kaynex Aug 25 '17 at 13:34
  • Yes. I meant the modulus of a complex function. This reference says that |z| is differentiable only at z = 0: https://rutherglen.science.mq.edu.au/wchen/lnicafolder/ica03.pdf

    Example 3.3.1. Consider the function f(z) = |z|^2 In our usual notation, we clearly have: u = x^2 + y^2 and v = 0. The Cauchy-Riemann equations 2x = 0 and 2y = 0 can only be satisfied at z = 0. It follows that the function is differentiable only at the point z = 0, and is therefore analytic nowhere. Is this incorrect?

     – Hass Aug 25 '17 at 13:39
  • As per the argument above, the Cauchy Riemann equations hold (only) for z = 0 – Hass Aug 25 '17 at 13:54
  • Hass, you argue that if the Cauchy-Riemann equations are satisfied at a point then the function is complex differentiable at said point. Beware that this is a false statement. – zokomoko Aug 25 '17 at 18:49

2 Answers2

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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.

Mark Viola
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  • It makes sense to use the limit definition of the derivative. So it's correct to work with the real derivative of |f(z)|, but not its complex derivative which does not exist. – Hass Aug 25 '17 at 14:03
  • So if f(z) is analytic, is |f(z)|^2 complex differentiable at U=V=0 , without it being analytic? If so is the derivative at U=V=0 equal to zero? – Hass Aug 26 '17 at 19:32
  • If $f$ is analytic, then $|f|^2$ is differentiable at any point $z$ such that $f(z)=0$ and its derivative is $0$. – Mark Viola Aug 26 '17 at 19:37
  • A formal proof or reference would be helpful. My take is that if f(z) = 0, then |f(z)|^2 is at its minimum, thus its real derivative = 0 which implies complex differentiation with a derivative = 0 – Hass Aug 26 '17 at 19:49
  • I gave a proof. Note that if $f(z)=0$, then $\overline{f(z)}=0$ and $$\lim_{\Delta z\to 0}\frac{\text{Re}(\overline{f(z)}f'(z)\Delta z)}{\Delta z}=0$$ – Mark Viola Aug 26 '17 at 20:28
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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)

zokomoko
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