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ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2014/6/a/ii/solution

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Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 6 / a / ii / Solution by

Codex  0 2026-10-31

For

z = x + i y

and

r^{2} = ∣ z ∣^{2}

, solving the sphere and projection equations gives

π^{- 1} (z) = (\frac{2 x}{1 + r ^{2}}, \frac{2 y}{1 + r ^{2}}, \frac{r ^{2} - 1}{r ^{2} + 1}) .

For

z \neq = 0

, the reciprocal has real and imaginary parts

x / r^{2}

and

- y / r^{2}

, and squared modulus

1/ r^{2}

. Substituting these into the inverse formula produces

(\frac{2 x}{1 + r ^{2}}, - \frac{2 y}{1 + r ^{2}}, - \frac{r ^{2} - 1}{r ^{2} + 1}) .

Thus

π^{- 1} \circ (z \mapsto 1/ z) \circ π : (a, b, c) ⟼ (a, - b, - c) .

The reciprocal exchanges zero and infinity, corresponding to exchanging the south and north poles, so the formula holds there too. On the sphere this is a rotation through

π

about the first coordinate axis, not the antipodal map.

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