 Solution

ID: past-exam-of-the-complex-analysis-conformal-maps-and-geometry-course-of-the-university-of-oxford/2023/1/b/iv/solution

Because

∣ h^{2} ∣ = ∣ h ∣^{2}

M (r) := \frac{1}{2 π} \int_{0}^{2 π} ∣ h^{2} (r e^{i θ}) ∣ d θ = n \geq 1 \sum ∣ c_{n} ∣^{2} r^{2 n} .

Differentiate this power series for

r < 1

and use iii:

M^{'} (r) = \frac{2}{r} \sum_{n \geq 1} n ∣ c_{n} ∣^{2} r^{2 n} = \frac{2 Area ( h ( r D ))}{π r} \leq \frac{2 r}{( 1 - r ^{2} ) ^{2}} .

Since

M (0) = 0

M (r) \leq \int_{0}^{r} \frac{2 s}{( 1 - s ^{2} ) ^{2}} d s = \frac{r ^{2}}{1 - r ^{2}} .

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