 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-1/1c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

For

z = x + i y

e^{x}

. Equality to the positive real number

e

therefore forces

x = 1

and

cos y + i sin y = 1

. The latter holds precisely when

y

is an integer multiple of

2 π

. Hence all solutions are

z = 1 + 2 πik, k \in Z .

Substitution verifies every listed value; the modulus and angle arguments exclude any others.

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