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ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2023/6/b/i/solution

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Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 6 / b / i / Solution by

Codex  0 2026-10-25

For

z = x + i y \in H

, direct calculation gives

∣ f (z) ∣^{2} = \frac{x ^{2} + ( y - 1 ) ^{2}}{x ^{2} + ( y + 1 ) ^{2}} < 1.

Conversely, solving

w = (z - i) / (z + i)

gives

z = i \frac{1 + w}{1 - w}, Im z = \frac{1 - ∣ w ∣ ^{2}}{∣1 - w ∣ ^{2}} > 0 (∣ w ∣ < 1) .

The denominator cannot vanish inside the unit disc, so these inverse formulas prove

f (H) = D

. This is the Cayley transform between the half-plane and disk.

For

z \in U

, we also require

∣ z ∣ > 1

. In inverse coordinates this becomes

\frac{∣1 + w ∣}{∣1 - w ∣} > 1 ⟺ ∣1 + w ∣^{2} - ∣1 - w ∣^{2} = 4 Re w > 0.

The equivalence holds in both directions, so

f (U) = {w \in C : ∣ w ∣ < 1, Re w > 0} .

Thus the exterior of the unit semicircle maps to the right half of the unit disc. The circular arc maps to its imaginary diameter, which is excluded because the source inequality is strict.

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