Let $B$ be <planar Brownian motion> started at $z\in D$, let $\tau_D$ be its exit time, and let $\phi:D\to D'$ be conformal. Define
$$
C_s=\int_0^{s\wedge\tau_D}|\phi'(B_r)|^2\,dr,
\qquad
\sigma_t=\inf\{s:C_s>t\}.
$$
The <conformal invariance of planar Brownian motion> states that
$$
\widetilde B_t=\phi(B_{\sigma_t}),
\qquad 0\leq t<C_{\tau_D},
$$
is Brownian motion started at $\phi(z)$ and stopped when it exits $D'$. Thus conformal maps preserve Brownian paths after this random time-change.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 203 / 2 / a / Solution

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
$$
p(z,e^{i\theta})
=\frac2\pi\frac{\operatorname{Im}z}{|z|^2}
\sin\theta\bigl(1+O(|z|^{-1})\bigr).
$$

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

If $B$ is planar <Brownian motion> in a domain $D$ and $f:D\to D'$ is a nonconstant <holomorphic function>, then $f(B)$, run with the clock $\int_0^t|f'(B_s)|^2ds$, is planar Brownian motion in $D'$ until the relevant exit time. In particular, <conformal maps> transport <harmonic measure> between planar domains.

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