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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-1/1b/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 1B / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

Write

z = x + i y

. Using the complex conjugate, the squared complex modulus on the left is

(α + β)^{2} x^{2} + (α - β)^{2} y^{2}

. Squaring the equality and dividing by

(α - β)^{2}

, which is nonzero, gives

x^{2} + y^{2} - 4 α β x - (α - β)^{2} = 0.

Completing the square identifies the candidate circle:

(x - 2 α β)^{2} + y^{2} = (α + β)^{2} .

Its centre is $2 α β$ on the real axis and its radius is $α + β$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle,

x \geq 2 α β - (α + β)

, so the original right-hand side satisfies

2 α β x + (α - β)^{2} \geq (α + β) (α + β - 2 α β) > 0.

Thus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

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