Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 13D / Solution

Let $v_1,v_2\ne 0$ be tangent vectors to the two smooth curves at $p$. complex differentiability at a point givesWhen $f^{\prime }(p)\ne 0$, its real derivative is multiplication by a nonzero complex number, which is a rotation followed by a positive scaling. It therefore preserves the angle between $v_1$ and $v_2$, so $f$ is conformal at $p$. The condition is essential: $f(z)=z^2$ has $f^{\prime }(0)=0$ and sends rays making angle $\pi /4$ at zero to rays making angle $\pi /2$.

For $J(z)=z+z^{-1}$,If $z,w$ both satisfy $|z|>1$, or both satisfy $0<|z|<1$, the alternative $zw=1$ is impossible unless the points lie on the omitted unit circle. Hence $J$ is one-to-one on each region. Also $J^{\prime }(z)=1-z^{-2}$ has no zero there, so the restrictions are conformal. Solving $z^2-wz+1=0$ shows that $w\in [-2,2]$ exactly when both roots lie on the unit circle; otherwise one root is inside and the other outside. Thus each region has image

Taking the reciprocal of the interior restriction and filling its removable value at zero givesThe reciprocal sends $\mathbb{C}\setminus [-2,2]$ ontoand $g(0)=0$. Therefore $g$ is the required one-to-one conformal map from the unit disc.