For the vertical slit $A=(1,1+i\epsilon]$, choose the square-root branch asymptotic to $z-1$ at infinity. The normalized map is
$$
\psi_A(z)=\sqrt{(z-1)^2+\epsilon^2}+\sqrt{1+\epsilon^2},
$$
for which $\psi_A(0)=0$ and
$$
\psi_A'(0)=\frac1{\sqrt{1+\epsilon^2}}.
$$
Therefore
$$
\boxed{\mathbb P\bigl(\gamma[0,\infty)\cap(1,1+i\epsilon]
=\varnothing\bigr)
=(1+\epsilon^2)^{-5/16}.}
$$


Solution

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

 Solution

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