Let \(f = u+iv\) be complex-differentiable with continuous partial derivatives at a point \(z = re^{i\theta}\) with \(r\neq 0\). Show that \[ \dd{u}{r} = {1\over r}\dd{v}{\theta} \qquad \dd{v}{r} = -{1\over r}\dd{u}{\theta} .\]
Solution
With \(x=r\cos\theta\) and \(y=r\sin\theta\), the chain rule gives \[ u_r=u_x\cos\theta+u_y\sin\theta, \qquad u_\theta=-r u_x\sin\theta+r u_y\cos\theta, \] and similarly \[ v_r=v_x\cos\theta+v_y\sin\theta, \qquad v_\theta=-r v_x\sin\theta+r v_y\cos\theta. \] Using the Cauchy–Riemann equations \(u_x=v_y\) and \(u_y=-v_x\), \[ v_\theta =r u_y\sin\theta+r u_x\cos\theta =r u_r, \] so \[ u_r=\frac1r v_\theta. \] Likewise, \[ u_\theta =-r v_y\sin\theta-r v_x\cos\theta =-r v_r, \] and therefore \[ v_r=-\frac1r u_\theta. \]