 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/7/v/solution

Assume the image of the domain of

f

lies in the domain where

g

is holomorphic. The complex chain rule gives

(g \circ f)^{'} (z) = g^{'} (f (z)) f^{'} (z),

so the composition is holomorphic, including at points where

f^{'} = 0

S (x, y) = s (u (x, y), v (x, y))

Δ S = s_{uu} ∣\nabla u ∣^{2} + 2 s_{uv} \nabla u \cdot \nabla v + s_{vv} ∣\nabla v ∣^{2} + s_{u} Δ u + s_{v} Δ v .

∣\nabla u ∣^{2} = ∣\nabla v ∣^{2}

; parts (i) and (ii) remove the last three terms. Since

s

is harmonic in its own variables,

Δ S = ∣\nabla u ∣^{2} (s_{uu} + s_{vv}) = 0.

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