... that open set being $$\mathbb{C} - \{x\in\mathbb{R} : x\leq-\frac{1}{4}\}$$ and the boundary conditions being $f(0) = 0$ and $f'(0) = 1$.
Here is my first try and only idea so far:
- $Ci\frac{1 - z}{1 + z}$, $C > 0$, takes $\mathbb{D}$ to the upper half-plane
- $Ci\frac{1 - z}{1 + z}\rightarrow D\big[Ci\frac{1 - z}{1 + z}\big]^2 = -C\big[\frac{1 - z}{1 + z}\big]^2$ takes the upper half-plane to $\mathbb{C} - \{x : x\geq 0\}$
- $-C\big[\frac{1 - z}{1 + z}\big]^2\rightarrow C\big[\frac{1 - z}{1 + z}\big]^2 - \frac{1}{4}$ takes the slit complex plane to $\mathbb{C} - \{x\in\mathbb{R} : x\leq-\frac{1}{4}\}$
The only possible choice for constant $C$ is $C = \frac{1}{4}$ but that choice causes the condition on the derivative to not be satisfied.
The Koebe map $f(z) = \frac{z}{(1-z)^2}$ satisfies the conditions. It can be seen that $f$ maps onto the desired domain by seeing it as the following composition of functions:
$$ u(z) = \frac{1+z}{1-z}, \;\;\; g(z) = u^2(z),\;\; h(z) = \frac{1}{4}(g(z) - 1). $$
It is not difficult to check that $u$ maps the unit disk onto the half plane with $Re z >0$, $g$ maps this half plane onto $\mathbb{C}\setminus\{x\in \mathbb{R} : x \leq 0\}$, and $h$ maps this domain onto the desired slit domain. It is also easy to see that all of these maps are conformal, thus
$f(z) = \frac{z}{(1-z)^2} = \frac{(1+z)^2 - (1-z)^2}{4(1-z)^2} = \frac14\left[ \left(\frac{1+z}{1-z} \right)^2 - 1\right]$
is also a conformal mapping. To check the differentiability condition, it is easy to calculate using elementary methods that the Taylor series expansion of $f$ is $\sum_{n=1}^\infty n z^n$, from which it follows that $f'(0) = 1$. (or of course calculate $f'(z)$ in the usual way and plug in $0$.)