0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ If someone can explain the down vote it would be great! $\endgroup$ – mathcounterexamples.net Sep 21 '16 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.