Let $Q = \{(x,y): x,y\geq 0\} $ be the 1st quadrant of $\mathbb R^2$, and $f$ is a function defined on it such that all the partial derivative(any order) of $f$ exists and continuous. By Whitney extension theorem (1934 Proceedings of AMS) we know there exists a functions $\tilde{f}$ and open set $\tilde{U}$, such that $\tilde{f}$ is defined on $\tilde{U}$ and $\tilde{f}$ is smooth, $Q\subset \tilde{U}$ and $\tilde{f}$ restricted to $Q$ is $f$. Also $\tilde{f}$ is not unique but for each $p\in Q$, $d\tilde{f}\_p$ is same. Now define $df\_p:= d\tilde{f}\_p$


Now for each $p\in Q$ , let $R(p)$ be a constant time rotation matrix at $p$, This mean 
$R(p)= c(p)
\left[ {\begin{array}{cc}
 \cos \theta(p) & \sin \theta(p),\\
-\sin \theta(p) & \cos \theta(p) 
 \end{array} } \right]$ 


$R(p)$ is $2\times 2$ matrix (pls someone fix the tex), and $c(p)$ is differentiable function on $Q$ to $\mathbb R$.
and we have for each $p\in Q$, $df\_p= R(p)$, does there exits any $\tilde{f}$ and $\tilde{U}$ as above such that for each $q\in \tilde{U}$ we have $d\tilde{f}\_q= S(q)$. Where $S(q)$ is constant time rotational matrix and $S(p)= R(p)$ for each $p\in Q$. 



Conformal Extension from a closed set to open
Ah, so you just mean that, when you regard $\mathbb{R}^2$ as $\mathbb{C}$ and you have a complex-valued function $f$ on $Q$, the closed first quadrant of $\mathbb{C}$, that satisfies the Cauchy-Riemann equations up to and including the boundary of $Q$, then does it extend holomorphically across the boundary. (I've never heard of your matrices $R(p)$ being called 'constant time rotation matrices. Where did you get that name?)


Well, the answer is 'no'. Using the Riemann Mapping Theorem, one can construct examples of such $f$ that are not real-analytic on the boundary of $Q$, which implies that $f$ does not extend holomorphically across the boundary, which is what you are asking for. 


The corner is not really relevant. You can even assume that $f$ is defined and smooth on the closed half-plane $x\ge0$ and construct examples such that $f$ does not extend across the line $x=0$ at any point. It suffices to consider a domain such as $x\ge g(y)$ where $g:\mathbb{R}\to\mathbb{R}$ is, say, smooth and bounded, but nowhere real-analytic. Then let $f$ biholomorphically map the domain $x>0$ onto the domain $x>g(y)$ and carry $\infty$ to $\infty$. Then $f$ will extend smoothly to a mapping that carries the line $x=0$ to the curve $x=g(y)$, but $f$ won't be real-analytic anywhere along that line, so it can't extend holomorphically to any domain that contains it.


