= Solution
On the parabolic domain with the nonpositive real axis removed, first apply the <principal square root>, then scale by $\pi/2$, and then apply the map of part (b). This gives the stated composite $C(z)$ and maps bijectively onto the slit <unit disc>.
There is no genuine <branch cut> in the composite. The function $\tan^2(\pi s/4)$ is even in $s$, so changing the local square root has no effect. Near zero its <Taylor series> starts as
$$
\tan^2(\pi\sqrt z/4)=\frac{\pi^2}{16}z+O(z^2),
$$
giving a <holomorphic> extension with $C(0)=0$. There are no <poles> within the parabolic domain: the possible squared locations are $z=(2+4k)^2$, all on the excluded real portion $x\geq4$.
For $z=-r^2$ with $r\geq0$, the continuation gives
$$
C(-r^2)=-\tanh^2(\pi r/4).
$$
As $r$ ranges from zero to infinity, these values run once through $(-1,0]$. They fill precisely the slit omitted by the previous mapping, and no point away from the negative real axis has an image in that slit. Therefore \b[the extended map is a bijection onto the whole unit disc]:
$$
\boxed{C:\{x+iy:y^2<4(1-x)\}\longrightarrow\{w:|w|<1\}.}
$$
It is <holomorphic> throughout, so this bijection is the desired <conformal map from a parabolic domain to the unit disc>; in particular $C'(0)=\pi^2/16\ne0$.
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