A harmonic function on a open connected set is analytic? My question is simple and short. According to Conway, if $u:G\to \mathbb{R}$ ($G\subset \mathbb{C}$) is harmonic, then it is infinitely differentiable. Does it mean that $u$ is analytic? 
 A: Since the discussion in the comments cleared up some points of terminology for me, I'll provide an answer to what seems like an implicit question in the question. 
On $\mathbb{C}$, there are several notions of analytic. Recall that the idea of an analytic function is that our function is equal, in some domain, to a convergent power series. Power series are particular kinds of limits of polynomials, and the question of real analytic vs. complex analytic is of what coordinate functions / variables we write our polynomials in. 
There are the real coordinate functions, x and y, and the complex coordinate z. Power series written in x and y are real analytic, and power series written in $z$ are complex analytic. (Power series written in $\bar{z}$ are anti-complex analytic, I guess.)
Writing power series in $z, \bar{z}$ is the same as writing it in $x,y$, if we allow our polynomials to have complex coefficients. This is because $z = x + iy$ and $\bar{z} = x - iy$.
The linked answer shows that any harmonic function $u$ is the real part of a holomorphic function $f = u + iv$. Thus, if we write our holomorphic function locally as $u + iv = \Sigma (a_n + i b_n) (x + i y)^n$, the real part is a power series (in $x$ and $y$), which starts as $a_0 + a_1x - b_1y + a_2(x^2 - y^2) - b_2xy + \ldots$. This is "real-analytic", as in Lord Shark the Unknown's comment.
In general, if we have a vector space $V$ over a field $K$, then the $K$-analytic functions are those which are power series over the $K$-linear coordinate functions on $V$. (Note that $x$ on $\mathbb{C}$ is not $\mathbb{C}$ linear, since $x(i) = 0$, while $x(1) = 1$.) So if $L \subset K$ is a subfield, then the notion of $L$-analytic function includes more functions, since if a functional is $K$ linear, then it is in particular $L$ linear. In our case, taking $\mathbb{R} \subset \mathbb{C}$, we recover the distinction between real or complex analytic on $\mathbb{C}^n$.
