Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 13A / b / Solution

It helps to separate the tangent and the squaring map. Put $t=\tan (z/2)$, with $z=x+iy$. The identityshows that $0<x<\pi /2$ gives $\mathrm{Re}t>0$ and $|t|<1$. To see that every point of this right half-disc occurs, setThen $z=-i\mathrm{Log}R$ has $0<\mathrm{Re}z<\pi /2$ and satisfies $\tan (z/2)=t$. The principal complex logarithm is unambiguous because $R$ is in the first quadrant.

Squaring maps this right half-disc bijectively onto the unit disc with the interval $(-1,0]$ removed: the argument of $t$ lies between $-\pi /2$ and $\pi /2$, and doubles into $(-\pi ,\pi )$. Thus the desired image isNeither derivative vanishes in the domain, so the composition is also a conformal map onto that slit disc.