Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 13A (Complex Analysis or Complex Methods)

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.

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.

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 asgiving 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 $x\ge 4$.

For $z=-r^2$ with $r\ge 0$, the continuation givesAs $r$ 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 the extended map is a bijection onto the whole unit disc:It is holomorphic throughout, so this bijection is the desired conformal map from a parabolic domain to the unit disc; in particular $C^{\prime }(0)={\pi }^2/16\ne 0$.