Thus the Pólya field rotates the usual representation of (f) by reflecting across the real axis. Write (f(z) = u + i v). Then:
The field ((v, u)) appears as the Pólya field of (-i f(z)). Connection to harmonic functions Since (f) is analytic, (u) and (v) are harmonic and satisfy the Cauchy–Riemann equations:
Thus (\nabla \psi = (v, u)). Check integrability: (\partial_x (v) = v_x = u_y) and (\partial_y (u) = u_y) — they match. So (\psi) exists (since domain simply connected). So:
[ u_x = v_y, \quad u_y = -v_x. ]
