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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 30D / b / ii / Solution

At a saddle the numerator of

ϕ^{'}

vanishes, so differentiating the quotient there gives

ϕ^{''} (z_{\pm}) = \frac{2 ( z _{\pm} - t )}{( z _{\pm}^{2} - 1 ) ( z _{\pm} - t )} = \frac{2}{z _{\pm}^{2} - 1} .

Since

z_{+}^{2} - 1 = 2 sin θ e^{i (θ + π /2)}, z_{-}^{2} - 1 = 2 sin θ e^{- i (θ + π /2)},

we obtain

ar g ϕ^{''} (z_{\pm}) = \mp (θ + π /2) .

The tangent angle

α

must satisfy

ar g ϕ^{''} (z_{\pm}) + 2 α = π (mod 2 π) .

An anticlockwise deformation of the original contour through both saddles can be oriented with

α_{+} = \frac{θ}{2} + \frac{3 π}{4}, α_{-} = \frac{π}{4} - \frac{θ}{2} .

Angles differing by

π

describe the same unoriented tangent; the displayed choices give the orientation needed to pass through the lower saddle and then the upper saddle while still enclosing

t

.

Solved by gpt-5.6-sol high.

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