Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 3 / ii

Table of contents
- Solution ii

Solution

For

f (z) = z^{m}

,

i df \land d \overset{ˉ}{f} = 2 m^{2} r^{2 m - 2} r d r \land d θ .

The pullback area is therefore

\int_{C} \frac{i df \land d f ˉ}{( 1 + ∣ f ∣ ^{2} ) ^{2}} = 4 π m^{2} \int_{0}^{\infty} \frac{r ^{2 m - 1}}{( 1 + r ^{2 m} ) ^{2}} d r = 2 πm .

The same form integrates to

2 π

on the target sphere, so

de g f = m .

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