 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2010/analysis/4/iii/solution

Write

f = u + i v

and evaluate all derivatives of

ψ

(u, v)

. The real chain rule gives

Δ (ψ \circ f) = ψ_{uu} (u_{x}^{2} + u_{y}^{2}) + 2 ψ_{uv} (u_{x} v_{x} + u_{y} v_{y}) + ψ_{vv} (v_{x}^{2} + v_{y}^{2}) + ψ_{u} Δ u + ψ_{v} Δ v .

u_{x}^{2} + u_{y}^{2} = v_{x}^{2} + v_{y}^{2} = ∣ f^{'} ∣^{2}

and zero cross term

u_{x} v_{x} + u_{y} v_{y} = 0

. Part (ii) removes the last two terms. Therefore

Δ (ψ \circ f) = ∣ f^{'} ∣^{2} (Δ ψ) \circ f = 0.

The composition is

C^{2}

, so it is harmonic. No assumption that

f^{'}

is nonzero is required.

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