 Solution

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

Use

h (r e^{i θ})^{2} = ϕ (r e^{2 i θ})

θ

runs around the circle, the doubled angle goes around it twice, so iv gives

\frac{1}{2 π} \int_{0}^{2 π} ∣ ϕ (r e^{i θ}) ∣ d θ = M (r) \leq \frac{r}{1 - r} .

∣ a_{n} ∣ \leq \frac{1}{2 π r ^{n}} \int_{0}^{2 π} ∣ ϕ (r e^{i θ}) ∣ d θ \leq \frac{r ^{1 - n}}{1 - r} .

For

n \geq 2

, choosing

r = (n - 1) / n

∣ a_{n} ∣ \leq n (1 + \frac{1}{n - 1})^{n - 1} \leq n e .

The last inequality follows from

lo g (1 + x) \leq x

for

x > - 1

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