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ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/k/k3/c/solution

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Past exam of the options II course of the University of Oxford / 2012 / K / K3 / c / Solution by

Codex  0 2026-10-31

The complex velocity obtained from part (b) is

w^{'} (z) = - \frac{A}{( z - b ) ^{2}} + \frac{B}{( z - c ) ^{2}} .

For a nonzero planar potential doublet,

B / A = (a^{2} / b^{2}) e^{- 2 i α}

. Its zeros therefore satisfy

\frac{z - c}{z - b} = t_{\pm}, t_{\pm} = \pm \frac{a}{b} e^{- i α},

and hence

z_{\pm} = \frac{c - b t _{\pm}}{1 - t _{\pm}} = a \frac{a / b \mp e ^{- i α}}{1 \mp ( a / b ) e ^{- i α}} .

The denominators are nonzero because

a / b < 1

. For either sign,

∣ a / b \mp e^{- i α} ∣^{2} = ∣1 \mp (a / b) e^{- i α} ∣^{2},

so

∣ z_{\pm} ∣ = a

. They are distinct because

t_{+} \neq = t_{-}

and

t \mapsto (c - b t) / (1 - t)

is injective, its determinant being

c - b \neq = 0

. Clearing the velocity denominators gives a quadratic with these two roots, so there are no further finite stagnation points. Thus both stagnation points lie on the obstacle boundary.

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