Harmonic function; is arg(z) harmonic

Harmonic function

A real function φ(x, y) is harmonic in a region if it has continuous second-order partial derivatives there and satisfies Laplace's equation

∂²φ/∂x² + ∂²φ/∂y² = 0.

Is V = arg z harmonic?

V = arg z = tan⁻¹(y/x) (defined for z ≠ 0, on a branch).

V_x = −y/(x² + y²)           V_y = x/(x² + y²)
V_xx = 2xy/(x² + y²)²        V_yy = −2xy/(x² + y²)²

V_xx + V_yy = 0, so yes, V = arg z is harmonic (in any region not containing the origin and not crossing the branch cut).

Harmonic conjugate

We want U such that f = V + iU is analytic. The C–R equations require V_x = U_y and V_y = −U_x.

U_y = V_x = −y/(x² + y²)   →  U = −½ log(x² + y²) + g(x)
U_x = −V_y = −x/(x² + y²)  →  g′(x) = 0

So U = −½ log(x² + y²) = −log|z| (plus a constant).

Check: V + iU = θ − i ln r = −i(ln r + iθ) = −i log z, which is analytic.

Answer: Yes, harmonic. The conjugate is U = −log√(x² + y²) = −ln r (+ constant).

(If V is instead taken as the imaginary part of an analytic function, the function is log z = ln r + iθ, and the corresponding real part is u = ln r.)