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I've learnd that, given an open $\mathbb{\Omega}$, if a function $f : \mathbb{\Omega} \subset \mathbb{C} \longrightarrow \mathbb{C} $ is holomorphic in some open $A \subset \mathbb{\Omega}$ the Cauchy-Riemann equations hold in $A$. So the contrapositive would be helpful to show that a function is not holomorphic meaning that if the function doesn't hold Cauchy-Riemann equations in an open $A \subset \mathbb{\Omega}$ then certainly the function is not holomorphic in $A$ (is everything ok until here?).

But I am quite stuck with this example:

$$f : \mathbb{C} \longrightarrow \mathbb{C},\quad f(z) = |z|^2 $$

$f$ is not holomorphic by definition beacuse it is only differentiable on $0$ and nowhere else. But my problem came out when I try to apply Cauchy-Riemann equations and I get that they only hold at the point $z = 0$. But ${0}$ is not open so, can I make some conclusion of $f$ about being or not holomorphic?? am I misunderstanding something??

Any help would be really appreciated!!

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  • $\begingroup$ Yes. At a point, complex differentiable means real differentiable and the jacobian satisfies the Cauchy Riemann equations. Holomorphic means complex differentiable on some open. $\endgroup$
    – reuns
    Oct 9, 2018 at 0:50
  • $\begingroup$ Yes but ... Suppose I was given the problem and I start attacking it by using Cauchy -Riemann equations and I just get that theese equations only hold at 0, What can I do from here?? $\endgroup$
    – Aldebaran
    Oct 9, 2018 at 0:59
  • $\begingroup$ $f(z) = |z|^2$ is real differentiable everywhere (that is for every $a,b \in \mathbb{C}$, $\mathbb{R} \ni t \mapsto f(a+tb)$ has a derivative) and the C-R equation holds only at $z=0$ $\endgroup$
    – reuns
    Oct 9, 2018 at 1:54

1 Answer 1

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You are correct in saying that the function $$f : \mathbb{C} \longrightarrow \mathbb{C},\quad f(z) = |z|^2$$

is not holomorphic at zero because there is not an open neighborhood of zero on which the function is differentiable.

As you mentioned , to be differentiable at one point is not sufficient for that function to be holomorphic at that point.

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  • $\begingroup$ Yes, but... Suppose I was given the problem and I start attacking it by using Cauchy -Riemann equations and I just get that theese equations only hold at $0$, What can I do from here?? $\endgroup$
    – Aldebaran
    Oct 9, 2018 at 0:56
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    $\begingroup$ The function is differentiable at $z=0$ but it is not analytic or holomorphic at that point. $\endgroup$ Oct 9, 2018 at 1:04

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