 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/1c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

The powers of the imaginary unit have period four. Hence

\sum_{a = 0}^{200} i^{a} = 50 (1 + i - 1 - i) + i^{200} = 1,

while

\prod_{b = 1}^{50} i^{b} = i^{1 + 2 + \dots + 50} = i^{1275} = i^{3} = - i .

Thus

x + i y = 1 - i

, so

x y = - 1 .

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