# Too complex to be true but too symmetrical to be false

For $i\in\{1,2\}$, let $D_i$ be a complex domain and let $z_i \in D_i$. Let $B_i$ be a complex Brownian motion starting from $z_i$. Set $T_i = \inf \{t > 0 : B_i(t) \notin D_i\}$.

Suppose that there exists a conformal isomorphism $\varphi : D_1 \to D_2$ such that $\varphi (z_1) = z_2$ . Set $\tilde{T} = \int_0^T |\varphi' (B_1(t))|dt$ and define for $t< \tilde{T}$

$\tau(t)= \inf \{s>0 : \int_0^s|\varphi' (B_1(r))|dr=t\}$ and $\tilde{B}(t) = \varphi(B_1(\tau(t)))$.

Then $(\tilde{T}, (\tilde{B}(t))_{t<\tilde{T}})$ and $(T_2, (B_2(t))_{t have the same distribution.