Milne-Thompson Theorem, I don't understand the terms ,nor the proof The Milne-Thompson theorem stats that if $f(x)$ is the complex potential of a flow with no rigid boundaries, no singularities inside $|z|=a$, then when introducing the solid cylinder $|z|=a$, the complex potential is $$w(z) = f(z)+ \bar{f}\left(\frac{a^2}{z}\right) $$ for $|z| \geq a$.

what does $\bar{f}\left(\frac{a^2}{z}\right) $ even mean? I know what a complex conjugate is, so if $z= a+bi$ then $\bar{z} = a-bi$. But what does this means?
Also, often they write also $\bar{f(\frac{a^2}{z}) }$. Is it the same thing then?
finally when proving it, they all prove that $|z|=a$ is a streamline and that the singularities outside the cylinder are the same. But how is this a proof of the formula??

 A: We are considering here two-dimensional potential flow, which is the irrotational flow (no vorticity) of an inviscid fluid (no viscosity).  For such flows the velocity field $(u,v)$ has zero curl and must be the gradient of a potential $\phi:$
$$\tag{*}u = \frac{\partial \phi}{\partial x}, \,\,\,\,\, v = \frac{\partial \phi}{\partial y}$$
The velocity components can also be related to a stream function $\psi$ by
$$\tag{**}u = \frac{\partial \psi}{\partial y}, \,\,\,\,\, v = -\frac{\partial \psi}{\partial x}$$
Any velocity field that satisfies (*) and (**) is automatically a solution of the Euler and continuity equations for fluid motion since $\nabla^2 \phi = 0$.  Thus, the specification of the correct flow for a given geometry reduces to ensuring that the proper boundary conditions are satisfied.  For an inviscid fluid, the velocity component normal to a solid surface (e.g., a cylinder) must be zero.  This is equivalent to ensuring that the surface is a streamline -- a curve where $\psi$ equals a constant.  This follows from observing that the dot product of the velocity field and the gradient of the stream function is zero at a point on a curve of constant $\psi:$
$$(u,v) \cdot \nabla \psi = \frac{\partial \psi}{\partial y}\frac{\partial \psi}{\partial x} +  \left(-\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial y}=  0$$
You bring up  "... they all prove that $|z| = a$ is a streamline," and raise the question " ... how is this a proof ..."  This, in fact, is exactly what needs to be shown in order to verify that the Milne-Thomson theorem provides a solution to flow around a cylinder.
A complex potential $f$ incorporates a velocity potential and stream function by $f = \phi + i \psi.$  By virtue of the Cauchy-Riemann equations at points where $f$ is analytic, the velocity components appear as
$$\frac{df}{dz} = u - iv$$
By showing that the combination of complex potentials has a vanishing imaginary part at the surface of the cylinder where $|z| = a$, we prove that the stream function is constant on the surface and the necessary normal velocity boundary condition is satisfied. 
To clarify the conjugate notation, consider a  complex-valued function $F(z) = U(x,y) + i V(x,y)$ of the complex variable $z = x + iy.$ An overbar over the entire expression $F(z)$, which is a complex number, means
$$\overline{F(z)} = U(x,y) - iV(x,y),$$
and $\overline{F}$ is defined to be the function with value
$$\overline{F}(z) = \overline{F(\overline{z})} = U(x,-y) - iV(x,-y)$$
Finally, the proof should now be clear. If $z = r e^{i \theta}$ is a point outside of the cylinder with $|z| > a$, then $a^2/z = (a^2/r)e^{-i \theta}$ is a point inside the cylinder with $|z| < a$. Any singularities of the second term now lie within the cylinder and do not affect the external velocity field. Furthermore, if $z$ is on the surface, then $z = ae^{i \theta}$, $a^2/z = ae^{-i \theta}$, and
$$\overline{z} = \frac{a^2}{z}, \,\,\,\,\, (\text{if} \,\,|z| = a).$$
Thus, for $|z| = a$, we have
$$\begin{align}f(z) + \overline{f}\left(\frac{a^2}{z}\right) &= f(z) + \overline{f\left(\frac{a^2}{\overline{z}}\right)} \\ &= f(z) + \overline{f\left(\overline{\overline{z}}\right)} \\  &= f(z) + \overline{f(z)} \\ &= 2 \,\Re{f(z)}\end{align}$$
Consequently the imaginary part (stream function) takes on a constant value of zero at the surface.
