Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 13A / a / 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 becomesThus the image is the strip $\boxed{0<\mathrm{Re}w<1}$. 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.