Here's my proof attempt:

$$f(z) = |z|^2$$

When $z=x+yi$, we have:

$$f(z) = x^2+y^2 + 0i$$

Doing Cauchy Riemann

$$\frac{\partial u}{\partial x} = 2x = \frac{\partial v}{\partial y} = 0$$

only when $x=0$. The second Cauchy Riemann condition is

$$\frac{\partial v}{\partial x} = 0 = -\frac{\partial u}{\partial y} = 2y$$

only when $y = 0$.

Therefore, the function can be differentiable only on $(0,0)$ but we haven't proven it yet, we must show that the partial derivatives are continuous on $(0,0)$, but they are. Therefore, $f(z)$ is only complex differentiable on $(0,0)$

Is everything all right?

  • $\begingroup$ When $z=x+yi$, then $|z|=x^2+y^2+0i$ and $f(z)=(x^2+y^2)^2$. $\endgroup$ – scott Sep 21 '16 at 20:40
  • $\begingroup$ @scott oops, you're rigth, but the reasoning is the same, right? $\endgroup$ – Guerlando OCs Sep 21 '16 at 20:42
  • $\begingroup$ @scott ?? no ${}{}{}$ $\endgroup$ – reuns Sep 21 '16 at 20:42
  • $\begingroup$ I would say instead that as $z \to 0$ : $f(z) = \mathcal{O}(|z|^2) = o(z)$ so it is complex differentiable. But as $z \to z_0\ne 0$ : $f(z) - f(z_0) = (|z| +|z_0|)(|z|-|z_0|) \ne C( z-z_0) + o(|z-z_0|)$ so it is not complex differentiable. $\endgroup$ – reuns Sep 21 '16 at 20:44
  • $\begingroup$ @scott wait, no... $\endgroup$ – Guerlando OCs Sep 21 '16 at 21:04

It looks good except when you say "we must show that the partial derivatives are continuous". It would be enough to prove that $f$ is differentiable as a real function of two variables at the origin.

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    $\begingroup$ If someone can explain the down vote it would be great! $\endgroup$ – mathcounterexamples.net Sep 21 '16 at 20:46

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