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!!
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.