Let $A=\mathbb H\cap B(1,\epsilon)$. Its normalized mapping-out map is
$$
\psi_A(z)=z+\frac{\epsilon^2}{z-1}+\epsilon^2,
$$
so $\psi_A(0)=0$ and
$$
\psi_A'(0)=1-\epsilon^2.
$$
Using the restriction exponent $5/8$ gives
$$
\boxed{\mathbb P\bigl(\gamma[0,\infty)\cap
(\mathbb H\cap B(1,\epsilon))=\varnothing\bigr)
=(1-\epsilon^2)^{5/8}.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 4 / d / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-203/4/d/i/solution