Let $f (z) = u+iv$ be an analytic function, then show that $ (∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2 $.

Question

Let $f (z) = u+iv$ be an analytic function, then show that $$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2\,.$$

$f(z) = u + iv $

$ϕ = |f(z)|^2 = u^2 + v^2 $

$f'(z) = ∂u/∂x + i∂v/∂x $

$|f'(z)|^2 = (∂u/∂x)^2 + (∂v/∂x)^2 = (∂u/∂y)^2 + (∂v/∂y)^2 $

$∂ϕ/∂x = 2u∂u/∂x + 2v∂v/∂x $

$∂^2ϕ/∂x^2 = (∂u/∂x)^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + (∂v/∂x)^2 $

$∂^2ϕ/∂x^2 = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) $ ----(1)

Similarly for,

$∂^2ϕ/∂y^2 = 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $ ----(2)

(1)+(2),

$(∂^2ϕ/∂x^2 + ∂^2ϕ/∂y^2) = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $

$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $

How it that the extra terms remain ??

Answer 1

Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$\frac {\partial ^{2}} {\partial x^{2}} u+ \frac {\partial ^{2}} {\partial y^{2}}u=0$ and $\frac {\partial ^{2}} {\partial x^{2}} v+ \frac {\partial ^{2}} {\partial y^{2}}v=0$. You can prove this easily using C-R equations]. Hence the 'extra terms' just add up to $0$.