Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 3 / a / Solution

Since $A$ is a bounded operator, its exponential power series converges in operator norm. The identity $A^2=I$ gives $A^{2k}=I$ and $A^{2k+1}=A$. Separating the even and odd powers therefore givesThe algebraic identity does not require $A$ to be Hermitian. Hermiticity is additionally needed for this exponential to be a unitary operator for real $\theta$.