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

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- Solution b

Solution

Put

F (z) = z + z^{- 1}

. On the unit semicircle,

F (e^{i θ}) = 2 cos θ

, and

\frac{d}{d θ} F (e^{i θ}) = 2 sin θ .

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 θ

:

p (z, e^{i θ}) = \frac{1}{π} \frac{Im F ( z )}{∣ F ( z ) - 2 c o s θ ∣ ^{2}} 2 sin θ .

z \to \infty

in

H

,

Im F (z) = Im z (1 - \frac{1}{∣ z ∣ ^{2}}), ∣ F (z) - 2 cos θ ∣^{2} = ∣ z ∣^{2} (1 + O (∣ z ∣^{- 1})),

uniformly in

θ \in [0, π]

. Hence

p (z, e^{i θ}) = \frac{2}{π} \frac{Im z}{∣ z ∣ ^{2}} sin θ (1 + O (∣ z ∣^{- 1})) .

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