Compute the hyperbolic metric on \[ \Delta^*=\Delta\setminus\{0\}. \]
Solution
The hyperbolic metric is \[ \boxed{ \rho_{\Delta^*}(z) =\frac{|dz|}{|z|\log(1/|z|)} =\frac{|dz|}{|z|\,|\log|z||}.} \]
1. The map\[ \pi(\zeta)=e^{2\pi i\zeta} \]is a holomorphic covering \(\mathbb H\to\Delta^*\).
Write \(\zeta=x+iy\) with \(y>0\). Then \[ |\pi(\zeta)|=e^{-2\pi y}\in(0,1), \] so the image lies in \(\Delta^*\). Conversely, if \(0<|z|<1\), choose an argument \(\theta\) of \(z\) and set \[ \zeta=\frac{\theta}{2\pi} +i\frac{\log(1/|z|)}{2\pi}. \] Then \(\zeta\in\mathbb H\) and \(\pi(\zeta)=z\). The deck transformations are \(\zeta\mapsto\zeta+k\), \(k\in\mathbb Z\), so \(\pi\) is the usual universal covering of the punctured disk.
2. If \(z=\pi(\zeta)\) and \(\zeta=x+iy\), then\[ y=\frac{\log(1/|z|)}{2\pi} \qquad\text{and}\qquad |\pi'(\zeta)|=2\pi|z|. \]
The first identity follows from \(|z|=e^{-2\pi y}\). The second follows from \[ \pi'(\zeta)=2\pi i e^{2\pi i\zeta}=2\pi i z. \]
3. The displayed metric is exactly the metric whose pullback under\(\pi\) is the hyperbolic metric \(|d\zeta|/\operatorname{Im}\zeta\) on\(\mathbb H\).
Let the metric on \(\Delta^*\) be \(\lambda(z)|dz|\). Since a universal covering between hyperbolic Riemann surfaces is a local isometry, it must satisfy \[ \lambda(\pi(\zeta))|\pi'(\zeta)|\,|d\zeta| =\frac{|d\zeta|}{y}. \] Using 2 gives \[ \lambda(z)\,2\pi|z| =\frac{2\pi}{\log(1/|z|)}. \] Hence \[ \lambda(z)=\frac{1}{|z|\log(1/|z|)}. \] Because \(0<|z|<1\), one has \(\log(1/|z|)=|\log|z||\), giving the equivalent form stated above.