 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-1/12/12-2a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For distinct

z, w

in the convex disc, integrate along the segment:

f (z) - f (w) = f^{'} (z_{0}) (z - w) + \int_{w}^{z} (f^{'} (ζ) - f^{'} (z_{0})) d ζ .

The second term has modulus strictly below

∣ f^{'} (z_{0}) ∣∣ z - w ∣

, so the sum cannot vanish. Thus

f

is one-to-one.

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