 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-1/3/b/iii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

i

are

i (π /2 + 2 πk)

with

k \in Z

, so

i^{1 - i} = exp [(1 + i) (π /2 + 2 πk)] .

Since the trigonometric factor is always

i

, all values in Cartesian form are

z = 0 + i e^{π /2 + 2 πk}, k \in Z .

k = 0

i e^{π /2}

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