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 these equations to show that the logarithm function defined by \[\log z = \log r + i \theta \; \; \mbox{where} \; z = r e^{i \theta } \; \mbox{with} \; - \pi < \theta < \pi\] is a holomorphic function in the region \(r>0, \; - \pi < \theta < \pi\). Also show that \(\log z\) defined above is not continuous in \(r>0\).
Solution
The beginning of the card is truncated. The missing part asks for the polar Cauchy–Riemann equations. If \(x=r\cos\theta\), \(y=r\sin\theta\), then the chain rule together with \(u_x=v_y\) and \(u_y=-v_x\) gives \[ u_r={1\over r}v_\theta, \qquad v_r=-{1\over r}u_\theta. \]
For \[ \Log z=\log r+i\theta, \qquad r>0,\quad -\pi<\theta<\pi, \] we have \(u_r=1/r\), \(u_\theta=0\), \(v_r=0\), and \(v_\theta=1\), so the polar Cauchy–Riemann equations hold. Hence the principal logarithm is holomorphic on the slit plane \(\CC\setminus(-\infty,0]\).
The final sentence of the source is false as literally written: a holomorphic function is continuous on its domain. The correct noncontinuity phenomenon is that this branch cannot extend continuously to all of \(\CC^\times\). Indeed, approaching \(-1\) from above gives imaginary part tending to \(\pi\), while approaching from below gives imaginary part tending to \(-\pi\). Thus no continuous extension across the negative real axis exists.