 Solution

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

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

z = e^{i θ}

. Factoring out the half-angle gives

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

Since

0 \leq θ < π

, the cosine is positive. Therefore

∣1 + z ∣ = 2 cos \frac{θ}{2}, ar g (1 + z) = \frac{θ}{2} .

On the Argand diagram, the points

0, 1, z, 1 + z

form a rhombus. Its diagonal from

0

1 + z

bisects the angle between the unit vectors

1

and

z

, so its direction is

θ /2

; resolving either side along the diagonal gives its length

2 cos (θ /2)

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