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<T_2}) have the same distribution.

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Too complex to be true but too symmetrical to be false

It’s a matter of principles (of convergence)

Let (f_n)_{n\in \mathbb{N}} be a sequence of entire functions such that, for every z \in \mathbb{C}, there exists F(z)=sup_{n\in \mathbb{N}}|f_n(z)| \in \mathbb{R}. Suppose that F:\mathbb{C} \to \mathbb{R} is continuous and F(0)=0. Let g be an entire function such that g'(0) \neq 0. Prove that g \circ f_n \to g over compact sets if and only if f_n(z) \to z, for every z \in \mathbb{C}.

It’s a matter of principles (of convergence)