Let [ f(z) = u(x,y) + i,v(x,y) ] be an analytic function on a domain (D \subset \mathbbC).
Thus the Pólya field rotates the usual representation of (f) by reflecting across the real axis. Write (f(z) = u + i v). Then:
So (\mathbfV_f) is (solenoidal) — it has a stream function. polya vector field
[ \mathbfV_f = (u,, -v). ]
Indeed, the stream function (\psi) such that (\mathbfV_f = ( \psi_y, -\psi_x )) can be taken as (\psi = -v). Check: [ \psi_y = -v_y = -(-u_x) = u_x? \text Wait carefully. ] Better: Let (\psi = -v). Then (\nabla^\perp \psi = (\psi_y, -\psi_x) = (-v_y, v_x)). But by Cauchy–Riemann, (v_x = u_y), (v_y = -u_x), so ((-v_y, v_x) = (u_x, u_y)) — that’s (\nabla u), not (\mathbfV_f). So that’s not correct. Let's derive cleanly: Let [ f(z) = u(x,y) + i,v(x,y) ]
[ u_x = v_y, \quad u_y = -v_x. ]
Equivalently, if (f = u+iv), then (\mathbfV_f = (u, -v)). The Pólya vector field is the conjugate of the complex velocity field (\overlinef(z)). Indeed, (\overlinef(z) = u - i v), which as a vector in (\mathbbR^2) is ((u, -v)). Then: So (\mathbfV_f) is (solenoidal) — it has
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:
[ \nabla u = (u_x, u_y) = (v_y, -v_x). ]
The of (f) is defined as the vector field in the plane given by
Radioplayer Canada is partnership between CBC Radio, the stations of Acadia Broadcasting, Bayshore Broadcasting, Blackgold Radio, Byrnes Communications, Central Ontario Broadcasting, Cogeco Media, Corus Radio, Durham Radio, Golden West Broadcasting, Pattison Media, Rawlco Radio, Rogers Media, RNC Media, Saskatoon Media Group, Stingray Radio, and Vista Radio, as well as the National Campus and Community Radio Association (NCRA/ANREC), among others.
