 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/30d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\frac{( z ^{2} - 1 ) ^{n}}{( z - t ) ^{n + 1}} = \frac{1}{z - t} (\frac{z ^{2} - 1}{z - t})^{n} .

Thus one may take

f (z, t) = \frac{1}{z - t}, ϕ (z, t) = lo g (\frac{z ^{2} - 1}{z - t}),

with a branch of the logarithm chosen locally along the deformed contour. Differentiation gives

ϕ^{'} (z, t) = \frac{2 z}{z ^{2} - 1} - \frac{1}{z - t} = \frac{z ^{2} - 2 t z + 1}{( z ^{2} - 1 ) ( z - t )} .

For

t = cos θ

z^{2} - 2 (cos θ) z + 1 = 0,

and hence

z_{\pm} = cos θ \pm i sin θ = e^{\pm i θ} .

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