Put $F(z)=z+z^{-1}$. On the unit semicircle, $F(e^{i\theta})=2\cos\theta$, and
$$
\left|\frac d{d\theta}F(e^{i\theta})\right|=2\sin\theta.
$$
The <conformal invariance of planar Brownian motion> and the <Poisson kernel for the upper half-plane> therefore give the exit density with respect to $d\theta$:
$$
p(z,e^{i\theta})
=\frac1\pi
\frac{\operatorname{Im}F(z)}
{|F(z)-2\cos\theta|^2}\,2\sin\theta.
$$
As $z\to\infty$ in $\mathbb H$,
$$
\operatorname{Im}F(z)
=\operatorname{Im}z\left(1-\frac1{|z|^2}\right),
\qquad
|F(z)-2\cos\theta|^2
=|z|^2\bigl(1+O(|z|^{-1})\bigr),
$$
uniformly in $\theta\in[0,\pi]$. Hence
$$
p(z,e^{i\theta})
=\frac2\pi\frac{\operatorname{Im}z}{|z|^2}
\sin\theta\bigl(1+O(|z|^{-1})\bigr).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 203 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-203/1/b/solution