# 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.

# 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 $z \in \mathbb{C}$.

# Convergence is contagious

Suppose $g$ is function $g: \Bbb C\to \Bbb C$ and that at a point $z_0\in \Bbb C$ such function is analytic. If the infinite sum $g(z_0)+g'(z_0)+\cdots$ converges, then in fact $g$ is entire and such sum converges for any other choice of $w$.