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 Incoming links: Laplacian identity for the squared modulus of a holomorphic function

Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 1 / 3D / Solution Created 2026-09-24 Updated 2026-10-14

Write the holomorphic function as

f = u + i v

. Analyticity gives smooth real and imaginary parts satisfying the Cauchy-Riemann equations

u_{x} = v_{y}

,

u_{y} = - v_{x}

. Differentiating these identities shows

Δ u = v_{y x} - v_{x y} = 0, Δ v = - u_{y x} + u_{x y} = 0.

Thus both parts are harmonic functions. The product rule then gives

Δ∣ f ∣^{2} = Δ (u^{2} + v^{2}) = 2 (u_{x}^{2} + u_{y}^{2} + v_{x}^{2} + v_{y}^{2}) + 2 u Δ u + 2 v Δ v = 4 (u_{x}^{2} + v_{x}^{2}) .

Since

f^{'} = u_{x} + i v_{x}

, this proves the Laplacian identity for the squared modulus of a holomorphic function,

Δ∣ f (z) ∣^{2} = 4∣ f^{'} (z) ∣^{2} .

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