Holomorphic Extension of a Sobolev Function is Sobolev I would like to start this question with an example that I had in mind. Let $f(x) = (x^2+1)^{-1}$, then we know a few of facts about $f(x)$. Firstly, $f\in L_2(\mathbb{R})$ and secondly, we also know that $f$ is analytic. Thirdly, we can also say that $f\in W^{1,2}(\mathbb{R})$, where $W^{k,p}(\Omega)$ is a Sobolev Space where the weak partial derivatives up to order $k$ belong to $L_p(\Omega)$.
Then by complex analysis, we can extend $f$ to a holomophic function $g$, namely $g(z) = (z^2+1)^{-1}$ onto a complex strip with upper and lower boundaries of $z= -i+\delta,i-\delta$ for some small $\delta >0$. Furthermore, along this strip $g$ is also in $W^{1,2}(\mathbb{C})$ or if you'd like, holding either the real or imaginary part constant in $W^{1,2}(\mathbb{R})$.
My question is, if we have an analytic function $f$, that is in $W^{1,2}(\mathbb{R})$, can we always find a holomorphic extension $g$ so that holding either the imaginary or real part fixed, $g\in W^{1,2}(\mathbb{R})$? (This result is weaker than just saying $g\in W^{1,2}(\mathbb{C})$).
 A: 
can we always find a holomorphic extension $g$ 

Note that by the identity theorem such an extension is unique. (Maybe you knew this, but the language of this sentence suggests otherwise.)

holding either the imaginary or real part fixed, $g\in W^{1,2}(\mathbb R)$.

The example you gave has this property for fixed imaginary part $y$ only when $y\notin \pm 1$, and for constant real part $x$ only when $x\ne 0$. 
But it can get worse. Take 
$$f(x) = \sum_{n=1}^\infty \frac{c_n}{(x-z_n)(x-\overline z_n)} \tag1$$ 
where $z_n=n^2+i/n^2$ and the numbers $c_n>0$ are made sufficiently small so that the series (1) converges in $W^{1,2}(\mathbb R)$. The series (1) converges locally uniformly on an open set containing $\mathbb R$, namely $U=\{z:|\operatorname{Im}z|<\min(1,1/|\operatorname{Re}z|)\}$. Therefore, $f$ is real-analytic on $\mathbb R$. However, it does not have a holomorphic extension to any strip of the form $\{z:|\operatorname{Im}z|<\delta\}$, due to the presence of poles. 
And it can get worse still. As before, let  $U=\{z:|\operatorname{Im}z|<\min(1,1/|\operatorname{Re}z|)\}$. Let $\Phi$ be a conformal map of $U$ onto a Jordan domain bounded by a nowhere differentiable curve. Then $\Phi$ is bounded and analytic on $\mathbb R$, but does not have holomorphic or even meromorphic extension to any domain properly containing $U$. The function $\exp(-z^2)\Phi(z)$ is in $L^2(\mathbb R)\cap L^1(\mathbb R)$. Let $F$ be its antiderivative; note that $F$ is bounded on $\mathbb R$. Finally, let $f(z)=\exp(-z^2)F(z)$. It is easy to check that $f$ is analytic on $\mathbb R$ and belongs to $W^{1,2}(\mathbb R)$. However, it does not have a  meromorphic extension to any domain properly containing $U$. So, the question of integrals along lines is moot: we don't have a function to integrate along lines. 
