 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2/5/c/solution

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

2 i

gives

e^{i z} - e^{- i z} = - 2 (e^{i z} + e^{- i z})

, hence

3 e^{i z} + e^{- i z} = 0

. Since exponentials never vanish, this is equivalent to

e^{2 i z} = - 1/3

. All branches of the complex logarithm give

2 i z = - ln 3 + i (π + 2 πn), z = \frac{π}{2} + nπ + \frac{i}{2} ln 3, n \in Z .

The derivation does not divide by

cos z

, so it introduces no lost zero-denominator cases.

 New to topics? Read the docs here!