= Solution
Take the <principal square root> $w=u+iv$, which has $u>0$ on the slit domain. The inverse relation $z=w^2$ gives $x=u^2-v^2$ and $y=2uv$. Hence the defining inequality for the parabolic domain becomes
$$
y^2<4(1-x)\quad\Longleftrightarrow\quad (u^2-1)(1+v^2)<0\quad\Longleftrightarrow\quad u<1.
$$
Thus \b[the image is the strip] $\boxed{0<\operatorname{Re}w<1}$. Conversely, every point of this strip squares to a point of the parabolic domain outside the removed nonpositive real axis, and its <principal square root> is the original $w$. This proves that the map is onto the strip and one-to-one. Its derivative $1/(2\sqrt z)$ is nonzero there, so it is a <conformal map>.
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