Extension of real or imaginary part of entire function to complex valued functions In this forum post on MSE we were trying to find the imaginary part of an entire function given it's real parte. One of the answers called for this method $$f(z) = 2u\left({z\over 2},-{{iz}\over 2}\right)-u(0,0) \tag{1}$$ which I've never seen before. This caught a lot of attention for the fact that, based on it's definition $$u(x,y):\mathbb{R}^2\rightarrow\mathbb{R}$$ and by using equation $(1)$ we are implying that we can extend the domain of $u(x,y)$ from $\mathbb{R}^2$ to $\mathbb{C}^2$, mainly $$u(z,w):\mathbb{C}^2\rightarrow\mathbb{C}$$ As you can see in the forum post, this started some back and forth between some users, me as well, to try and find some mathematical rigour to equation $(1)$. Because of this I'm now here asking for that so desired mathematical proof: when and why is it possible to extend real-valued function to complex-valued functions? Moreover: when is it possible to apply equation $(1)$ to find an entire function from it's real part?
My approach
I copy here my attempt of an explanation and a sort of proof that I've given even in the comment of the MSE post cited: 
The real part of an entire function can be expressed as
$$2u(x,y) = f(x+iy)+\overline{f(x+iy)} = f(x+iy)+\overline{f(\overline{x-iy})} = f(x+iy)+\overline{f}(x-iy)\tag{2}$$
by choosing $$z=x+iy\;\;w=x-iy \Rightarrow x={1\over 2}(z+w)\;\;y={1\over{2i}}(z-w)$$ then $$2u\left({1\over 2}(z+w), {1\over{2i}}(z-w)\right) = f(z)+\overline{f}(w)= f(z)+\overline{f(\overline{w})}\;\;\forall z,w$$
So we can choose $w$ arbitrarily and set it to $w = z_0$ which for the problem given was $z_0=0+i0$, then $$f(z)=2u\left({1\over 2}(z+z_0), {1\over{2i}}(z-z_0)\right)-\overline{f(z_0)} \Rightarrow f(z)=2u\left({1\over 2}z, {1\over{2i}}z\right)-u(0,0)$$
This same proof can be used to find that $$f(z) = 2iv\left({z\over 2},{z\over{2i}}\right)+v(0,0)$$
The problem comes again when substituting in equation $(2)$ for the fact that $z,w\in\mathbb{C}$ and my answer was that the right hand side has sense for every $x$ and $y$,real or complex, such that $x+iy$ and $x-iy$ are in the domain of $f$ which clearly are being $f$ an entire function, so analytic an all $\mathbb{C}$.
Final thoughts
So in the end, I would like some comments by you! If it's possible, I'd love to have a clear and mathematical rigorous explanation.
 A: 
when and why is it possible to extend real-valued function to complex-valued functions?

is the wrong question, it's too general. We can't work with just any extension, we need certain regularity properties. We want a holomorphic extension. Technically, for $f$ defined by $(1)$ to be holomorphic it is not neccessary that $u$ is holomorphic, $z \mapsto u(z/2,-iz/2)$ can be holomorphic even if $(z,w) \mapsto u(z,w)$ isn't. But we can always work with a holomorphic extension.
The local version of the extension question becomes

Let $U \subset \mathbb{R}^2$ be open, and $u \colon U \to \mathbb{R}$. When does there exist an open $\Omega \subset \mathbb{C}^2$ with $U \subset\Omega \cap \mathbb{R}^2$ and a holomorphic $\tilde{u} \colon \Omega \to \mathbb{C}$ such that $\tilde{u}\rvert_{U} = u$?

The answer is "if and only if $u$ is real-analytic". That's clearly necessary, since the Taylor expansion of $\tilde{u}$ about a point in $U$ gives a power series expansion of $u$ about that point by restricting to real arguments. And it is sufficient because the Taylor series of $u$ about a point $p \in U$ converges on some neighbourhood $W_p$ of $p$ in $\mathbb{C}^2$  and thus provides a holomorphic extension of $u\rvert_{W_p \cap \mathbb{R}^2}$ to $W_p$. If we take appropriate $W_p$, say balls or polydisks with centre $p$, then all these extensions fit together and provide a holomorphic extension $\tilde{u}$ with domain
$$\bigcup_{p \in U} W_p\,.$$
A global version of the extension question is

Let $u \colon \mathbb{R}^2 \to \mathbb{R}$. When does there exist a holomorphic $\tilde{u} \colon \mathbb{C}^2 \to \mathbb{C}$ with $\tilde{u}\rvert_{\mathbb{R}^2} = u$?

The answer to this question is easy too. If such a $\tilde{u}$ exists, its Taylor series (about $0$, say) converges on all of $\mathbb{C}^2$, and thus $u$ has a globally convergent power series expansion. Conversely if $u$ has a power series expansion that converges on all of $\mathbb{R}^2$, this series converges on all of $\mathbb{C}^2$ and thus provides an entire $\tilde{u}$.
Now, if $u \colon \mathbb{R}^2 \to \mathbb{R}$ is real-analytic, it need not be the case that $u$ has a globally convergent series expansion. Take for example $u(x,y) = \frac{1}{1+x^2+y^2}$. Its extension is meromorphic with the nonempty pole set $\{ (z,w) \in \mathbb{C}^2 : z^2 + w^2 = -1\}$.
But, the $u$ we want to extend is not an arbitrary real-analytic function. It is supposed to be the real part of a holomorphic function, and hence it must be harmonic. [Which yields sanity check 1 for such problems: Is the given function harmonic? If it isn't, it cannot be the real (or imaginary) part of a holomorphic function.] And an entire harmonic function has a globally convergent power series representation. (Take the Poisson integral over a sphere of radius $R$ and expand the Poisson kernel into a power series. This shows that the Taylor series about $0$ converges at least on the ball of radius $R$.)
Thus when we're looking for an entire function with a prescribed real part - and the given function is harmonic - the method always works.
The method can fail if we want to find a holomorphic $f \colon \Omega \to \mathbb{C}$ with prescribed real part $u$ when $\Omega \subsetneq \mathbb{C}$, since $(z/2, -iz/2)$ need not belong to the domain of the holomorphic extension $\tilde{u}$ for $z\in \Omega$ in that case.
Ahlfors gives a short discussion of the method in section 1.2 of chapter 2 in his Complex Analysis.
