Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 35 / 2 / b / ii / Solution

Let ${\varphi }_i:D\to U_i$ be the normalized maps, and put $V={\varphi }_1(U_2)$. A simple loop in $U_1$ has its interior there, so applying ${\varphi }_1^{-1}$ preserves the property of surrounding $0$. There is a disjoint decompositionThe conformal restriction property identifies the measure of the second set with $\nu ({\mathbf{L}}_D\setminus {\mathbf{L}}_{U_2})$. ThereforeThe composition fixes $0$ and has positive derivative there, so it is exactly the normalized map onto $V$. No interchange of infinite differences is needed: the argument is additivity on disjoint sets. The later logarithmic expression is consistent with this identity because $({\varphi }_1\circ {\varphi }_2{)}^{\prime }(0)={\varphi }_1^{\prime }(0){\varphi }_2^{\prime }(0)$.