 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-3/13f/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

No. The nonvanishing holomorphic functions

f_{n} (z) = e^{n z}

satisfy

f_{n} (0) = 1

, but

∣ f_{n} (1/2) ∣ = e^{n /2} ⟶ \infty.

Avoiding zero alone permits an unbounded holomorphic logarithm's real part; the omitted ray in part (c) was what constrained the image after taking a square root.

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