Milne Thomson method for determining an analytic function from its real part What is the logic behind taking  $z = {\bar {z}}$ while finding analytic function in Milne-Thomson Method ? 
I mean we write 
$ { f(z)=u(x,y)+iv(x,y)} $ 
as 
$ 
{\displaystyle f(z)=u\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar {z}}}{2i}}\right)+iv\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar {z}}}{2i}}\right)} {\displaystyle } $
Its fine uptil now but then we say $f(z)$ can be regarded as an identity in two independent variables $z$ and $ {\bar {z}}$
so we take $z = {\bar {z}}$, this gives us
$ { f(z)=u(z,0)+iv(z,0)} $
Source  
I am not getting the the part $z = {\bar {z}}$ . How do we regard it an identity of 2 independent variable? Please guide me with an intuitive explanation. ( Please use example if you can ) .
I went through this Link  but could not understand how $w$ is independent of $z$. I mean whatever value of $x$ and $y$ we take $z$ and $w$ are obtained from the same x and y, so how can we say both are independent!
Thanks in advance !
 A: Milne Thomson's method allows to reconstruct an analytic function $f(z)$ from its real part $u(x,y)$, when the latter is given as an "analytic expression" in terms of $x$ and $y$. The presentation of this method in Wikipedia (and other sources)  is taken verbatim from Milne Thomson's original paper. Since it remained unclear to me why the method works I tried to prove the result without resorting to Wirtinger derivatives. Here goes:
For simplicity assume that $z\mapsto f(z)=u(x,y)+iv(x,y)$ is  analytic  in a disc  $\>U\colon\>|z|<\rho$. For $z\in U$ one has
$$f'(z)=\lim_{h\to0, \>h\in{\mathbb R}}{f(z+h)-f(z)\over h}={\partial f\over\partial x}(z)=u_x(x,y)+iv_x(x,y)\ ,\tag{0}$$  hence by the CR equations
$$f'(z)=u_x(x,y)-iu_y(x,y)\qquad(z\in U)\ .\tag{1}$$
There are real constants $a_k$, $b_k$ $(k\geq0)$ such that
$$f'(z)=\sum_{k=0}^\infty(a_k+ib_k)\>(x+iy)^k\qquad(z\in U)\ .$$
With $(1)$ it then follows that
$$\left.\eqalign{\phi(x)&:=u_x(x,0)={\rm Re} \big(f'(x)\bigr)=\sum_{k=0}^\infty a_k\>x^k\cr
\psi(x)&:=- u_y(x,0)={\rm Im} \big(f'(x)\bigr)=\sum_{k=0}^\infty b_k\>x^k\cr}\right\}\qquad(-\rho<x<\rho)\ .$$
These series representations imply that $\phi$ and $\psi$ are in fact restrictions of certain analytic functions 
 $$z\mapsto\phi(z),\quad z\mapsto\psi(z)\qquad(z\in U)$$ to the real axis. Now $(1)$ implies $$f'(x)=u_x(x,0)-i u_y(x,0)=\phi(x)+i\psi(x)\qquad(-\rho<x<\rho)\ ,$$
and this allows to conclude that one even has
$$f'(z)=\phi(z)+i\psi(z)\qquad(z\in U)\ .\tag{2}$$
Let $\Phi$ and $\Psi$ be analytic primitives of $\phi$ and $\psi$ in $U$. Then from $(2)$ it follows that
$$f(z)=\Phi(z)+i \Psi(z)+ c\ ,\tag{3}$$
whereby $c$ is defined up to an imaginary constant.
Example. Let $u(x,y):=\cos x\, e^{-y}$. Then
$$\phi(x)=u_x(x,0)=-\sin x,\qquad \psi(x)=-u_y(x,0)=\cos x\ .$$
It follows that $\phi(z)=-\sin z$, $\>\psi(z)=\cos z$, so that we may take $\Phi(z)=\cos z$, $\>\Psi(z)=\sin z$. Using $(3)$ we then obtain
$$f(z)=\cos z+i\sin z+c=e^{iz}+c\ ,$$
whereby $u(0,0)=1$ requires $c\in i{\mathbb R}$.
A: I know I'm a little late to the party but I think this will feel more intuitive.
\begin{align}
f(z) &= u(x,y) + \imath v(x,y) \\
f(x + \imath y) &= u(x,y) + \imath v(x,y)
\end{align}
Put $ y=0 $,
\[ f(x) = u(x, 0) + \imath v(x,0) \]
Now, $x$ becomes just a dummy variable for the function $f$. Replacing $x$ by $z$ we get,
\[ f(z) = u(z,0) + \imath v(z,0) \]
This exact thing is reflected when we take $ z=\bar{z} $, that is, we are putting $y=0$. We are first expressing the function using only a single real variable and then saying that the variable can also take on complex values.
