 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-1/1b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The exponent in the PDF is

2 z^{2}

. A complex exponential equals one precisely when its exponent is an integer multiple of

2 πi

: its real part must be zero, and its imaginary part must be a full rotation. Therefore

z^{2} = πik

for

k \in Z

. Taking both square roots yields all solutions

{0} \cup {\pm \frac{πk}{2} (1 + i) : k = 1, 2, \dots} \cup {\pm \frac{πk}{2} (1 - i) : k = 1, 2, \dots} .

The first nonzero family corresponds to positive exponent index, and the second to negative index. Both signs are required.

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