Is complex analytic extension of real-analytic diffeomorphism a diffeomorphism ? Hi, my question is :
Let $D$ be the open unit disk in $\mathbb{C}=R^2$ and $f:D\to D$ be a real-analytic diffeomorphism. Let us think of the canonical embedding : $\mathbb{C}=R^2\subset \mathbb{C^2}. $ Does there exist a complex analytic diffeomorphism $F$ ( analytic in two complex variables ) whose domain is either $D^2\subset \mathbb{C^2}$ or the complex 2-dimensional unit open polydisk $\Omega ={{(z,w): |z|^2+|w|^2 < 1}}$ in $\mathbb{C^2}$ such that its restriction to $D\subset D^2$ or $D\subset \Omega $ is $f$ ? By restriction , I mean $ F(z,0) = f(z) $ in the case of $ D \subset D^2 \subset \mathbb {C}^2 $ .
The range of $F$ does not necessarily have to be $D^2$ or $\Omega$, 
but it would be even better if they are !
If this is a very well-known result, you can cite a reference. 
Is the same result true in 1-dimension as well , i.e. replacing $D$ by $I\subset R$ and changing the complex-analytic/conformal diffeomorphism $F$ accordingly , i.e. asking that domain of $F$ is $I^2$ or $D$ with restriction  $ f $ ?
Thank you .
 A: Indeed, a real-analytic map can always be extended to some complex neighbourhood. 
The problem is, the neighbourhood may be very small. Consider, for example, 
the map
$$f:I\to I,\, I=[-1,1],$$
defined by 
$$f:x\mapsto x+\frac{a^3x(x-1)^2}{x^2+a}.$$
For small $a>0$ this is a diffeomorphism of $I$, but it cannot be extended 
very much due to the poles near $x=0$. The similar map (though a bit more contrived)
can be designed for a disc. So, the answer to the question is no.
Of course, all this is well known but what is a proper reference I cannot say.
P.S. This is an answer to the question as I understand it. There are some 
points I do not understand. $\Omega$, as it is defined, is a sphere. 
And, I hope, the restriction is not defined by $F(z,0)=(f(z),0)$: if it is, 
you can always take $F(z,w)=(f(z),w)$.
A: I'm not 100% sure I understand your question, so pardon me if I'm saying something unrelated.
It seems to me that you are asking whether a given real-analytic function $f \colon \mathbb{R}^2 \to \mathbb{R}^2$ (perhaps only defined on suitable open subsets), given by
$$f(x,y) = [ f_1(x,y), f_2(x,y) ]$$
is in fact complex-analytic, i.e. there is $F \colon \mathbb{C} \to \mathbb{C}$ such that
$$f(x,y) = F(x+iy) = u(x,y) + iv(x,y)\text{ .}$$
This is patently the case if and only if $u$ and $v$, hence $f_1$ and $f_2$, satisfy the Cauchy-Riemann equations, and thus any pair of real-analytic functions $f_i \colon \mathbb{R}^2 \to \mathbb{R}$, $i=1,2$, which do not satisfy CR furnish a counter-example.
On the other hand, what you may have wanted to ask, is whether a real-analytic function extends to a complex-analytic function, e.g. if for a real $f(x)$ there is a complex $F(z)$ such that $F(\Re{z}) = f(\Re{z})$. This can indeed always be done on some Stein neighbourhood, but you don't in general have control over the size of the neighbourhood.
