Using $d\varphi=h^{-1}dh$,
$$
d(he^{-\varphi})=e^{-\varphi}(dh-h,d\varphi)=0.
$$
The overlap $\mathbb C^*$ is connected, so $he^{-\varphi}=c$ for some $c\in\mathbb C^*$. Choose $C\in\mathbb C$ with $e^C=c$ and replace $\varphi$ by $\varphi+C$. Then $h=e^\varphi$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 / 2 / g / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-115/2/g/solution