Holomorphic and analytic functions.

No. "Holomorphic" means "complex differentiable in a neighborhood of a point", "analytic" means "Taylor series at a point converges to the function in a neighborhood of that point".

For real analysis the analogue for holomorphic would be "real differentiable in a neighborhood of a point", but this condition is far too weak to imply the latter. E.g. there are functions which are only once differentiable but not twice, and even infinitely differentiable functions differentiable everywhere yet analytic nowhere.

It's worth adding to the above answer that there's a theorem which says that "holomorphic" can be ambiguous between "differentiable once" and "differentiable infinitely often" (i.e. in complex analysis the above identity of holomorphic and analytic also means that complex-differentiable-in-a-neighborhood implies complex-smooth-on-that-neighborhood).

However there also exist functions which are $C^\infty(\mathbb R^n, \mathbb R)$ and are not analytic. The most common example is $$f(x) = \exp(-1/x)\text{ if } x > 0\text{ else } 0.$$By induction the derivatives of $\exp(-1/x)$ have the form $e^{-1/x} P_n(1/x)$ for polynomial $P_n$, since we start out with that form $P_0 = 1$ and the derivative gives a simple recursive formula to calculate the next polynomial $P_{n+1}(u) = u^2~P_n(u) + P_n'(u)$ for a new polynomial. But $\lim_{u\to\infty} e^{-u}P_n(u) = 0$ and hence $\lim_{x\to0^+}f^{(n)}(x) = 0$ for all $n$, so all derivatives are continuous at the origin and this is $C^\infty(\mathbb R, \mathbb R)$ at $x=0$. However it is not analytic at $x=0$ precisely because the Taylor series at $x=0$ is the zero function, and $f$ is not the zero function.