Let (\phi = u) (potential). 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:

[ \nabla u = (u_x, u_y) = (v_y, -v_x). ]

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:

Thus the Pólya field rotates the usual representation of (f) by reflecting across the real axis. Write (f(z) = u + i v). Then:

Let [ f(z) = u(x,y) + i,v(x,y) ] be an analytic function on a domain (D \subset \mathbbC).

The of (f) is defined as the vector field in the plane given by

[ \mathbfV_f = (u,, -v). ]

[ \mathbfV_f(x,y) = \big( u(x,y),, -v(x,y) \big). ]