Let \(f=u+iv\) be differentiable (i.e. \(f'(z)\) exists) with continuous partial derivatives at a point \(z=re^{i\theta}\), \(r\not= 0\). Show that \[\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta},\quad \frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}.\]
Solution
Since \(f(z)\) is complex differentiable at \(z\), the Cartesian Cauchy-Riemann equations hold: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \]
The polar coordinates \((r, \theta)\) are related to \((x, y)\) by: \[ x = r \cos \theta, \qquad y = r \sin \theta. \] By the multivariable chain rule: \[ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta, \] \[ \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta} = -\frac{\partial u}{\partial x} r \sin \theta + \frac{\partial u}{\partial y} r \cos \theta. \] Similarly, for \(v\): \[ \frac{\partial v}{\partial r} = \frac{\partial v}{\partial x} \cos \theta + \frac{\partial v}{\partial y} \sin \theta, \] \[ \frac{\partial v}{\partial \theta} = -\frac{\partial v}{\partial x} r \sin \theta + \frac{\partial v}{\partial y} r \cos \theta. \]
Now substitute the Cartesian Cauchy-Riemann equations (\(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\)) into the expression for \(\frac{1}{r}\frac{\partial v}{\partial \theta}\): \[ \frac{1}{r}\frac{\partial v}{\partial \theta} = -\frac{\partial v}{\partial x} \sin \theta + \frac{\partial v}{\partial y} \cos \theta = \frac{\partial u}{\partial y} \sin \theta + \frac{\partial u}{\partial x} \cos \theta = \frac{\partial u}{\partial r}. \] Similarly, substitute into \(-\frac{1}{r}\frac{\partial u}{\partial \theta}\): \[ -\frac{1}{r}\frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \sin \theta - \frac{\partial u}{\partial y} \cos \theta = \frac{\partial v}{\partial y} \sin \theta - \left(-\frac{\partial v}{\partial x}\right) \cos \theta = \frac{\partial v}{\partial x} \cos \theta + \frac{\partial v}{\partial y} \sin \theta = \frac{\partial v}{\partial r}. \]
Thus, the polar Cauchy-Riemann equations hold: \[ \frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta}, \qquad \frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}. \]