 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2007/b/8/ii/solution

Choose

r > 0

so the closed disk centered at

a

r

lies in

U

f (a) = \frac{1}{2 πi} \int_{∣ z - a ∣ = r} \frac{f ( z )}{z - a} d z .

Parametrize

z = a + r e^{i θ}

; then

d z = i r e^{i θ} d θ

and the denominator cancels. Thus

f (a) = \frac{1}{2 π} \int_{0}^{2 π} f (a + r e^{i θ}) d θ .

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