Show that in polar coordinates, the Cauchy-Riemann equations take the form \[ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \text { and } \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} .\]
Use (a) to show that the logarithm function, defined as \[ \Log z=\log r+i \theta \text { where } z=r e^{i \theta} \text { with }-\pi<\theta<\pi .\] is holomorphic on the region \(r> 0, -\pi < \theta < \pi\).
Show also that this branch cannot be extended continuously across the negative real axis to all of \(\CC\setminus\{0\}\).
Solution
For (a), write \(x=r\cos\theta\) and \(y=r\sin\theta\). Then \[ u_r&=u_x\cos\theta+u_y\sin\theta =v_y\cos\theta-v_x\sin\theta ={1\over r}v_\theta,\\ v_r&=v_x\cos\theta+v_y\sin\theta =-u_y\cos\theta+u_x\sin\theta =-{1\over r}u_\theta, \] using the Cartesian Cauchy–Riemann equations \(u_x=v_y\) and \(u_y=-v_x\).
For (b), take \[ u(r,\theta)=\log r, \qquad v(r,\theta)=\theta. \] Then \[ u_r={1\over r}={1\over r}v_\theta, \qquad v_r=0=-{1\over r}u_\theta, \] so the principal branch \(\Log z=\log r+i\theta\) is holomorphic on \[ \CC\setminus(-\infty,0] =\{re^{i\theta}:r>0,\ -\pi<\theta<\pi\}. \] In particular it is continuous on that domain.
It cannot be extended continuously across the negative real axis. Indeed, \[ z_k^+=e^{i(\pi-1/k)}, \qquad z_k^-=e^{-i(\pi-1/k)} \] both tend to \(-1\), while \[ \Log z_k^+=i(\pi-1/k)\to i\pi, \qquad \Log z_k^-=-i(\pi-1/k)\to-i\pi. \] The two one-sided limits disagree, so no continuous extension to \(-1\), and hence no continuous extension of this branch to all of \(\CC^\times\), exists.