 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ib/paper-4/4a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Parametrize the positively oriented circle by

z = r e^{i θ}

. The Cauchy derivative formula at the origin becomes

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

d \overset{z}{ˉ} = - i r e^{- i θ} d θ

gives

0 = - i r \int_{0}^{2 π} \overline{f (r e^{i θ})} e^{- i θ} d θ .

Add this zero integral to the derivative integral. Since

f + \overset{ˉ}{f} = 2 Re f = 2 u

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

The complex conjugation on the integrand and the reversed sign in

d \overset{z}{ˉ}

are both required; the circle's orientation itself has not been reversed.

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