 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/a/solution

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

The point

1 + i

2

π /4

2 π

. Taking complex square roots halves the argument and takes the square root of the modulus:

∣ z ∣ = 2^{1/4}, ar g z = \frac{π}{8} + kπ (mod 2 π), k = 0, 1.

The two roots are opposite points in the complex plane, explicitly

\pm [(2 + 1) /2 + i (2 - 1) /2]

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