Past exam of the complex analysis: conformal maps and geometry course of the University of Oxford / 2023 / 1 / b / v

Table of contents
- Solution 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} .

The Cauchy integral formula for the Taylor coefficients now implies

∣ 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

gives the linear coefficient bound for normalized univalent functions,

∣ 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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