 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-1/1b/b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

1 + ω = e^{iπ /3}

are

i (\frac{π}{3} + 2 πk), k \in Z .

The definition of complex exponentiation therefore gives all values as

exp [i (1 + ω) (\frac{π}{3} + 2 πk)] .

Since

1 + ω = \frac{1}{2} + i \frac{3}{2}

, these can be written

(- 1)^{k} e^{- 3 π (k + 1/6)} e^{iπ /6}, k \in Z .

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