# Cauchy - Riemann equation conclusion

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!!

• Yes. At a point, complex differentiable means real differentiable and the jacobian satisfies the Cauchy Riemann equations. Holomorphic means complex differentiable on some open. Oct 9, 2018 at 0:50
• 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?? Oct 9, 2018 at 0:59
• $f(z) = |z|^2$ 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 $z=0$ Oct 9, 2018 at 1:54

You are correct in saying that the function $$f : \mathbb{C} \longrightarrow \mathbb{C},\quad f(z) = |z|^2$$
• 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?? Oct 9, 2018 at 0:56
• The function is differentiable at $z=0$ but it is not analytic or holomorphic at that point. Oct 9, 2018 at 1:04