1
$\begingroup$

I'm looking for a complex function $f(z)$ such that it's differentiable at $z = z_0$ but isn't continuous in any neighborhood of $z = z_0$. I don't think differentiability in one point implies continuity in some neighborhood but I couldn't find a counterexample for that.

This question occurred to me when I encountered with the following theorem: enter image description here

$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

$f(x)=z^{2}$ for $z$ on the real axis and $f(z)=z^{3}$ everywhere else is such a function.

$\endgroup$
12
  • $\begingroup$ Could you show how this function satisfies those conditions, please? $\endgroup$
    – S.H.W
    Jan 5, 2020 at 11:54
  • $\begingroup$ The derivative at $0$ is $0$ since $\frac {z^{k}-0} z \to 0$ as $ z\to 0$ for $k=2$ as well as $k=3$. Any open disk around $0$ contains points $z$ with $z^{2} \neq z^{3}$. At any such a point the function is not continuous. @S.H.W $\endgroup$
    – Kavi Rama Murthy
    Jan 5, 2020 at 11:57
  • $\begingroup$ Does this function satisfy in Cauchy–Riemann equations? And in general is continuity in some neighborhood necessary for these equations? $\endgroup$
    – S.H.W
    Jan 5, 2020 at 12:00
  • $\begingroup$ The partial derivatives don 't even exist in any open disk around the origin. $\endgroup$
    – Kavi Rama Murthy
    Jan 5, 2020 at 12:05
  • $\begingroup$ What about origin? In that point is it true that $u_x = v_y$ and $u_y = -v_x$? $\endgroup$
    – S.H.W
    Jan 5, 2020 at 12:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .