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

Write $A_r=[-r,r]\times(0,1]$ and take $r\geq1$. By the <Brownian representation of half-plane capacity>,
$$
\operatorname{hcap}(A_r)
=\lim_{y\to\infty}y\,
\mathbb E_{iy}[\operatorname{Im}B_\tau].
$$
On hitting $A_r$, the exit height is at most one. Moreover, $A_r$ lies in the half-disc of radius $\sqrt{r^2+1}\leq2r$. The <harmonic measure> of that semicircle as viewed from $iy$ is $O(r/y)$: mapping its exterior to $\mathbb H$ by $z\mapsto z+(2r)^2/z$ reduces the estimate to the <Poisson kernel for the upper half-plane> on an interval of length $O(r)$. Consequently
$$
\mathbb E_{iy}[\operatorname{Im}B_\tau]\leq\frac{Cr}{y}
$$
for large $y$, and $\operatorname{hcap}(A_r)\leq Cr$. This is the <half-plane capacity of a low rectangle>.

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

{c}
{title2=$P_y(x)=\dfrac1\pi\dfrac{y}{x^2+y^2}$}

For the <complex upper half-plane>, the Poisson kernel is
$$
P_y(x-v)=\frac1\pi\frac{y}{(x-v)^2+y^2}.
$$
It is the density of <harmonic measure> on the real boundary as viewed from $x+iy$.

= Poisson kernel
{synonym}

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 Incoming links: Poisson kernel for the upper half-plane