 Solution

ID: past-exam-of-the-complex-analysis-conformal-maps-and-geometry-course-of-the-university-of-oxford/2023/3/b/i/solution

For

0 < ∣ z ∣ < R

w = R^{2} / \overset{z}{ˉ}

, which lies outside the closed disk. On

∣ ζ ∣ = R

, direct algebra gives the Poisson kernel identity

Re \frac{ζ + z}{ζ - z} = \frac{ζ}{ζ - z} - \frac{ζ}{ζ - w} .

Using

d θ = d ζ / (i ζ)

\frac{1}{2 πi} \oint_{∣ ζ ∣ = R} f (ζ) [\frac{1}{ζ - z} - \frac{1}{ζ - w}] d ζ = f (z) .

The first term is

f (z)

by the Cauchy integral formula; the second vanishes by the Cauchy integral theorem, since its pole lies outside the disk. At

z = 0

the kernel is one, and the same identity is the mean value property for holomorphic functions. Therefore

f (z) = \frac{1}{2 π} \int_{0}^{2 π} f (R e^{i θ}) Re \frac{R e ^{i θ} + z}{R e ^{i θ} - z} d θ (∣ z ∣ < R) .

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