Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 203 / 1 / c / Solution

No such conformal bijection exists. A prescribed point is a polar point for planar Brownian motion: a planar Brownian motion started away from zero hits zero with probability zero. For example, inside an annulus the probability of reaching radius $\epsilon$ before radius $R$ isby the harmonic function $\log |w|$ and the optional stopping theorem. For fixed $R>|z^{\prime }|$ this tends to zero as $\epsilon \downarrow 0$. Any finite-time visit to zero would occur before leaving some sufficiently large disc; the countable union over integer $R$ still has probability zero.

The exit time from $\mathbb{C}\setminus \{0\}$ is therefore infinite almost surely, contradicting part (b) if this were the image of a bounded domain. Equivalently, a hypothetical inverse $\psi :\mathbb{C}\setminus \{0\}\to D$ would be bounded and holomorphic. The Riemann removable singularity theorem extends it over zero, and the Liouville theorem makes it constant, contradicting bijectivity.