Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 8B / b / Solution

Because $(n\cdot \sigma {)}^2=I$, even and odd powers sum separately: $R=I\cos \theta +i(n\cdot \sigma )\sin \theta$. Then $R{R}^{†}=I$.

Solved by gpt-5.6-sol high.