# Differentiability versus analyticity domains for complex functions

I'm having some trouble understanding the differences between the concepts of differentiablity and analyticity domains of a complex function.

I know that when a complex function $$f(z)$$ have a complex derivative at a point $$z_0$$ then it is complex differentiable at $$z_0$$, i.e, $$f'(z_0)$$ exists. When we say that a complex function is analytic in a domain $$D$$, that means that $$f'(z)$$ exists at every point $$z\in D$$.

So, what is the difference between the differentiability domain and the analyticity domain?

If I'm able to find a domain where the complex derivative exists then that same domain wouldn't be the analyticity domain?.

I think that it's true if that domain is open. But what happens when $$f(z)$$ is just differentiable at one point, would it be analytic at that point? I think,no. Because the definition of analyticity requires a neighborhood where the function is analytic. Am I wrong?

I'll put an example:

Given this complex function: $$f(z)=\frac{2z+1}{z^2+1}$$

I know that the complex derivative does not exists at the points $$z=\pm i$$. So the differentiability domain is $$\mathbb{C}-\{i,-i\}$$. Wouldn't the analyticity damain be the same? Is $$\mathbb{C}-\{i,-i\}$$ an open set?

Any help is appreciated.

Maybe it helps to know the original meanings of the words analytic and differentiable. Let $$U\subseteq\mathbb C$$ be open. A function $$f:U\to\mathbb C$$ is called complex differentiable in $$z_0\in U$$ if the limit of the difference quotient at $$z_0$$ exists (so the classical idea behind differentiability). It is called analytic in $$z_0$$ if there exists an open neighborhood of $$z_0$$ on which $$f(z)$$ is identical to a power series centered at $$z_0$$. That is,
$$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$$
for all $$z$$ in said open neighborhood. Now it turns out that if $$f$$ is analytic in $$z_0$$ according to this definition, then it is automatically analytic on the entire neighborhood in which it agrees with the power series. So if $$f$$ is analytic in a point, we can always find an open set on which it is also analytic. So in practice, we're always interested in analyticity on open sets.
So to answer your specific questions: Your example function is in fact analytic on $$\mathbb C-\{\mathrm i,-\mathrm i\}$$. But I could imagine a function which is only complex differentiable in a single point, like you mentioned yourself. For instance, $$z\mapsto\vert z\vert^2$$ is only complex differentiable on $$\{0\}$$, so it's not analytic.