Q:

A function is analytic if and only if it is holomorphic

I am studying some complex analysis and I have found this result. In fact it isn't a single theorem (I am reading Cartan's "Elementary theory of analytic functions" and I haven't found such a theorem, but maybe I just didn't find it), but there are many results that should give us this equivalence. However, I was wondering about two things. 
The first is: is this equivalence "perfect"? I mean, could I exchange the terms "holomorphic" and "analytic" in theorems and various results without changing anything, or are there some hypothesis I should satisfy first? I am asking this because even though they should be equivalent, I always find theorems about (for instance) analytic prolongation that always talk about analytic functions, and there is no "holomorphic prolongation", if you understand what I mean. So I was not sure if this was simply a matter of tradition or if analytic functions are more general/poweerful than holomorphic ones.
The second one is: if I have a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ that is differentiable in an open connected set $U$ and in that open set it satisfies Cauchy-Riemann conditions, then it should be holomorphic and analytic, if viewed as a complex function. So, for the prolongation theorem, there should be only one function like that (that is, differentiable and that satisfes Cauchy-Riemann)on $U$. So, are you telling me that if I have a differentiable function satisfying Cauchy-Riemann in an open connected set, not only it is the only function (unless you add/subtract constants obviously) that does that, but (being analytic in $\mathbb{C}$) it is also $C^{\infty}$. It would be so incredible I can't believe it!

A:

The answer to your final question is affirmative. Yes, the equivalence is perfect. Yes, if a function $f\colon\Omega\longrightarrow\mathbb C$, where $\Omega$ is an open subset of $\mathbb C$, is differentiable, then $f\in C^\infty$. Yes, it is incredible.

