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.
 A: 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.
You may ask what this definition of analyticity has to do with the one you were provided. It turns out that if a complex function is complex differentiable on an open set, it automatically has a power series representation. And any function which has a power series representation is automatically complex differentiable. So analyticity (the power series version) on an open set is equivalent to complex differentiability on that set. And many authors now use analytic to denote complex differentiability on an open set, knowing that it's equivalent to the original meaning.
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.
