 Solution

ID: past-exam-of-the-brownian-motion-in-complex-analysis-course-of-the-university-of-oxford/2011/2/a/solution

Let

β (θ) = w - e^{i θ} = ∣ β (θ) ∣ e^{i g (θ)}

. It is never zero, since

∣ w ∣ < 1

e^{i θ} - w = - β (θ), 1 - \overline{w} e^{i θ} = - e^{i θ} \overline{β (θ)} .

Dividing these identities gives

f (e^{i θ}) = e^{- i θ} \frac{β ( θ )}{β ( θ )} = exp (i (- θ + 2 g (θ))) .

Any other continuous lift of the argument differs by an integer multiple of

2 π

, which leaves the exponential unchanged.

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