Question: Map the outside of the parabola $y^2 = 2px$ on the disc $|w|<1$ such that $z=0$ and $z=-\frac{p}{2}$ are mapped to $w=1$ and $w=0$.
Note that the mapping must be conformal. While haven't made much progress, I think splitting the question into the following is the best idea: Map the outside to $\mathfrak{R}(w)>0$ which can then be easily mapped to $|w|<1$ via the transformation $w=\frac{z-1}{z+1}$ which is evident geometrically. But I'm not sure how to proceed with the first step. An idea I had was to let:
$f(z) = (y^2 - 2px) + iv(x,y)$ and with a surjective $v$ onto the right-half plane could be an effective transformation. But if we consider the Cauchy-Riemann equations:
$\frac{\partial u}{\partial x} = -2p = \frac{\partial v}{\partial y}$
$-\frac{\partial u}{\partial y} = -2y = \frac{\partial v}{\partial x}$
These equations are impossible to satisfy. I don't have any further ideas though.
It may be helpful to include another similar problem I have solved, and this may indicate flaws in my method to help me solve more types of these problems.
This question is: Map the interior of the right-hand branch of the hyperbola $x^2-y^2=a^2$ to $|w|<1$ such that the focus corresponds to $w=0$ and the vertex corresponds to $w=-1$.
Note that $f(z) = (x^2 - y^2 - a^2) + 2ixy = z^2 - a^2$ maps the inside of the parabola to the right half-plane (considering only $\mathfrak{R}(z)>0$ so that the left-branch is not mapped anywhere).
Hence the transformation:
$w = \frac{z^2 - a^2 - 1}{z^2 - a^2 + 1}$ maps the inside of the right hyperbola to $|w|<1$.
Note that $w(a) = -1$ and $w(√(a^2+1)) = 0$ while $w(a√2) = \frac{a^2 - 1}{a^2 + 1}$
We now want a transformation that maps the unit disc to itself, such that $w'(-1) = -1$, $w'(\frac{a^2 - 1}{a^2 + 1}) = 0$
We find that $w'(z) = \frac{\frac{a^2 - 1}{a^2 + 1}-z}{\frac{a^2 - 1}{a^2 + 1}z - 1} = \frac{(a^2 - 1)-(a^2+1)z}{(a^2-1)z - (a^2+1)}$
Overall we thus have the transformation:
$w(z) = \frac{(a^2 - 1)-(a^2+1)\frac{z^2 - a^2 - 1}{z^2 - a^2 + 1}}{(a^2-1)\frac{z^2 - a^2 - 1}{z^2 - a^2 + 1} - (a^2+1)} = \frac{z^2-2a^2}{z^2+1}$, where $\mathfrak{R}(z)>0$, is hence the required transformation (if I simplified correctly).
Let us assume $p>0$.
Consider $z^2$. The image of $y = y_0$ is $v^2= 4 y_0^2 (y_0^2 + u)$.
(For $p<0$. We simply work with fixed $x$ instead of fixed $y$.)
Let $2p=4 y_0^2$. Then $f(z) = (z-p/2)^{1/2}$ is well defined outside of the parabola. (We may choose which branch we like. But let's say we choose the one that gives us positive real parts.) It maps the parabola to a horizontal line $y_0=i (p/2)^{1/2}$, and the the outside of the parabola to $\mathrm{Im}z > (p/2)^{1/2}$.
Furthermore, it maps $0$ to $i(p/2)^{1/2}$ and $-p/2$ to $i(p)^{1/2}$.
The rest is mapping a half-plane to the unit disc.
Below is an explanation where you got it wrong in your attempt.
You were assuming that a family of parabolas $y^2-2px = c$, with $c$ being 0, 1, 2, etc., are mapped into equally spaced vertical lines. But from the solution above, we can see it is a family of parabolas $y^2 - 2 p x = p^2$, with $p$ being 1, 2, 3, etc., that are mapped into (not equally spaced) parallel straight lines.