 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/1a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Similarly,

1 - z = e^{i θ /2} (e^{- i θ /2} - e^{i θ /2}) = - 2 i sin \frac{θ}{2} e^{i θ /2} = 2 sin \frac{θ}{2} e^{i (θ /2 - π /2)} .

Thus, for

0 < θ < π

∣1 - z ∣ = 2 sin \frac{θ}{2}, ar g (1 - z) = \frac{θ}{2} - \frac{π}{2} .

θ = 0

1 - z = 0

, whose complex argument is undefined. Geometrically,

∣1 - z ∣

is the chord joining the two unit-circle points

z

and

1

. The isosceles triangle with vertex angle

θ

has half-chord

sin (θ /2)

, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

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