Thus the Pólya field rotates the usual representation of (f) by reflecting across the real axis. Write (f(z) = u + i v). Then:
The Pólya field (\mathbfV_f) is exactly (w) — so it is a (gradient of a harmonic function, also curl-free and divergence-free locally). polya vector field
The of (f) is defined as the vector field in the plane given by Thus the Pólya field rotates the usual representation
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: The of (f) is defined as the vector
We want (\mathbfV_f = (u, -v) = (\partial \psi / \partial y,; -\partial \psi / \partial x)). From the first component: (\partial \psi / \partial y = u). From the second: (-\partial \psi / \partial x = -v \Rightarrow \partial \psi / \partial x = v).
Let (\phi = u) (potential). Then